Inverse Trig Derivatives

Inverse trig derivatives are the derivatives of inverse trigonometric functions. We have 6 inverse trigonometric functions that are the inverses of the 6 basic trigonometric functions. The inverse trig derivatives are defined only in the domain of the inverse trigonometric functions which are stated as follows:

Let us learn the derivatives of inverse trigonometric functions with detailed proof. Also, let us solve a few problems related to the inverse trig derivatives.

What are Inverse Trig Derivatives?

The inverse trig derivatives are the derivatives of the inverse trigonometric functions arcsin (or sin -1 ), arccos (or cos -1 ), arctan (or tan -1 ), etc. We use implicit differentiation to find the derivatives of the inverse trig function which we we explore in detail in the upcoming section. Here are the inverse trig derivatives :

  • The derivative of arcsin x is d/dx(arcsin x) = 1/√ 1-x² , when -1 < x < 1
  • The derivative of arccos x is d/dx(arccos x) = -1/√ 1-x² , when -1 < x < 1
  • The derivative of arctan x is d/dx(arctan x) = 1/(1+x²), for all x in R
  • The derivative of arccsc x is d/dx(arccsc x) = -1/(|x|√ x²-1 ), when x < -1 or x > 1
  • The derivative of arcsec x is d/dx(arcsec x) = 1/(|x|√ x²-1 ), when x < -1 or x > 1
  • The derivative of arccot x is d/dx(arccot x) = -1/(1+x²), for all x in R

Derivatives of Inverse Trig Functions

We know that another form of writing an inverse trig function, say arcsin x, is sin -1 x. The derivatives of inverse trig functions can be written in alternative notation as follows:

d/dx (sin -1 x) = 1/√ 1-x²

  • d/dx(cos -1 x) = -1/√ 1-x²
  • d/dx(tan -1 x) = 1/(1+x²)
  • d/dx(csc -1 x) = -1/(|x|√ x²-1 )
  • d/dx(sec -1 x) = 1/(|x|√ x²-1 )
  • d/dx(cot -1 x) = -1/(1+x²)

Inverse trig derivatives or derivatives of inverse trig functions

Let us derive each of these derivative formulas.

Inverse Trig Derivatives Proofs

In the previous section, we have already seen the formulas of derivatives of inverse trigonometric functions. If we observe them carefully, the derivatives neither include trigonometric functions nor include inverse trigonometric functions. Rather they include squares and square roots . Thus, it is difficult to memorize them unless we know how these formulas are derived. We use the process of implicit differentiation (which is the process of using the chain rule when the functions are implicitly defined) to derive the inverse trig derivatives.

Derivative of Arcsin

To find the derivative of arcsin x, let us assume that y = arcsin x. Then by the definition of inverse sine , sin y = x. Differentiating both sides with respect to x,

cos y (dy/dx) = 1

dy/dx = 1/cos y ... (1)

By one of the trigonometric identities , sin 2 y + cos 2 y = 1. From this, cos y = √ 1-sin²y = √ 1-x² .

Substituting this in (1),

dy/dx = 1/√ 1-x² (or)

d (arcsin x) / dx = 1/√ 1-x²

Thus, the derivative of arcsin x (or) sin -1 x (or) inverse sin x is 1/√ 1-x² .

Derivative of Arccos

To find the derivative of arccos x, let us assume that y = arccos x. Then by the definition of inverse cos, cos y = x. Differentiating both sides with respect to x,

-sin y (dy/dx) = 1

dy/dx = 1/(-sin y) = -1/sin y ... (1)

By one of the trigonometric identities, sin 2 y + cos 2 y = 1. From this, sin y = √ 1-cos²y = √ 1-x² .

dy/dx = -1/√ 1-x² (or)

d (arccos x) / dx = 1/√ 1-x²

Thus, the derivative of arccos x (or) cos -1 x (or) inverse cos x is 1/√ 1-x² .

Derivative of Arctan

To find the derivative of arctan x, let us assume that y = arctan x. Then by the definition of inverse tan , tan y = x. Differentiating both sides with respect to x,

sec 2 y (dy/dx) = 1

dy/dx = 1/(sec 2 y) ... (1)

By one of the trigonometric identities, sec 2 y - tan 2 y = 1. From this, sec 2 y = 1 + tan 2 y = 1 + x 2 .

dy/dx = 1 / (1 + x 2 ) (or)

d (arctan x) / dx = 1 / (1 + x 2 )

Thus, the derivative of arctan x (or) tan -1 x (or) inverse tan x is 1 / (1 + x 2 ).

Derivative of Arccsc

To find the derivative of arccsc x, let us assume that y = arccsc x. Then by the definition of inverse cosecant, csc y = x. Differentiating both sides with respect to x,

-csc y cot y (dy/dx) = 1

dy/dx = -1/(csc y cot y) ... (1)

By one of the trigonometric identities, csc 2 y - cot 2 y = 1. From this, cot 2 y = csc 2 y - 1 = x 2 - 1. Then cot y = √ x²-1 .

Also, we have csc y = x.

dy/dx = -1/(|x|√ x²-1 )(or)

d (arccsc x) / dx = -1/(|x|√ x²-1 ).

Here, we have written the absolute value sign around x instead of just writing x because if we observe the graph of csc -1 x, the slope of the tangent of this curve is always negative. So the derivative of csc -1 x must be always negative irrespective of the sign of x. That is why we always write the absolute value sign around x here.

absolute value sign derivative of inverse trig function of cosecant

Thus, the derivative of arccsc x (or) csc -1 x (or) inverse csc x is -1/(|x|√ x²-1 ).

Derivative of Arcsec

To find the derivative of arcsec x , let us assume that y = arcsec x. Then by the definition of inverse cosecant, sec y = x. Differentiating both sides with respect to x,

sec y tan y (dy/dx) = 1

dy/dx = 1/(sec y tan y) ... (1)

By one of the trigonometric identities, sec 2 y - tan 2 y = 1. From this, tan 2 y = sec 2 y - 1 = x 2 - 1. Then tan y = √ x²-1 .

Also, we have sec y = x.

dy/dx = 1/(|x|√ x²-1 ) (or)

d (arcsec x) / dx = 1/(|x|√ x²-1 ).

Here also we have used the absolute value sign as the graph of sec -1 x always has tangents with positive slopes and hence the derivative shouldn't be affected by the sign of x.

absolute value sign inverse trig derivative of secant

Thus, the derivative of arcsec x (or) sec -1 x (or) inverse sec x is 1/(|x|√ x²-1 ).

Derivative of Arccot

To find the derivative of arccot x, let us assume that y = arccot x. Then by the definition of inverse cot , cot y = x. Differentiating both sides with respect to x,

-csc 2 y (dy/dx) = 1

dy/dx = -1/(csc 2 y) ... (1)

By one of the trigonometric identities, csc 2 y - cot 2 y = 1. From this, csc 2 y = 1 + cot 2 y = 1 + x 2 .

dy/dx = -1 / (1 + x 2 ) (or)

d (arccot x) / dx = -1 / (1 + x 2 )

Thus, the derivative of arccot x (or) cot -1 x (or) inverse cot x is -1 / (1 + x 2 ).

Inverse Trig Derivatives and Integrals

Here is a table with derivatives and integrals of inverse trigonometric functions. This will help you to summarize and memorize the difference between the derivatives and integrals of inverse trig functions.

☛ Related Topics:

  • Differentiation of Trigonometric Functions
  • Differentiation
  • Derivatives Calculator

Examples on Inverse Trig Derivatives

Example 1: What is the derivative of sin -1 (2x 3 )?

By the inverse trig derivatives,

Using this and also the chain rule,

d/dx (sin -1 (2x 3 )) = 1/√ 1-(2x³)² d/dx(2x 3 )

= 1/√ 1-4x⁶ (6x 2 )

Answer: The derivative of sin -1 (2x 3 ) is (6x 2 )/√ 1-4x⁶.

Example 2: Find the derivative of sin -1 x + cos -1 x.

By the derivatives of inverse trig functions,

d/dx (cos -1 x) = -1/√ 1-x²

Thus, d/dx (sin -1 x + cos -1 x)

= 1/√ 1-x² - 1/√ 1-x²

Alternative Method:

By inverse trig formulas, we have sin -1 x + cos -1 x = π/2

Differentiating the above equation on both sides,

d/dx (sin -1 x + cos -1 x) = d/dx (π/2) = 0.

(This is because the derivative of a constant is 0)

Answer: The derivative of sin -1 x + cos -1 x is 0.

Example 3: What is the derivative of x tan -1 x?

By inverse trig derivative formulas,

By product rule,

d/dx(x tan -1 x) = x d/dx(tan -1 x) + tan -1 x d/dx(x)

= x/(1+x²) + tan -1 x

Answer: The derivative of x tan -1 x is x/(1+x²) + tan -1 x.

go to slide go to slide go to slide

derivatives of inverse trig functions worksheet with answers

Book a Free Trial Class

Practice Questions on Inverse Trig Derivatives

go to slide go to slide

FAQs on Inverse Trig Derivatives

What are the formulas of inverse trig derivatives.

The inverse trig derivatives are the derivatives of the inverse trigonometric functions. They can be derived using the formulas of inverse trig functions and differentiation techniques. The most used formulas are:

  • d/dx(sin -1 x) = 1/√ 1-x²

How to Memorize Inverse Trig Derivatives?

The derivatives of inverse trig functions are:

  • d/dx(csc -1 x) = -1/(|x|√(x²-1))
  • d/dx(sec -1 x) = 1/(|x|√(x²-1))

Here, the derivatives of sin -1 x and cos -1 x are negatives of each other; the derivatives of tan -1 x and cot -1 x are negatives of each other; and the derivatives of csc -1 x and sec -1 x are negatives of each other.

How to Find the Derivatives of Inverse Trigonometric Functions?

To find the derivatives of inverse trigonometric functions, we use implicit differentiation. For example, to find the derivative of sin -1 x, we assume that y = sin -1 x from which we get sin y = x. Differentiating both sides with respect to x, we get cos y dy/dx = 1. From this, dy/dx = 1/cos y = 1/√ 1-sin²y = 1/√ 1-x² . Like this, we can derive the derivatives of other inverse trigonometric functions.

What is the Derivative of Sin -1 4x Using Inverse Trig Derivatives?

We know that d/dx(sin -1 x) = 1/√ 1-x² . Using this and also applying the chain rule, d/dx(sin -1 x) = 4/√ 1-16x² .

Where Can I Find Inverse Trig Derivatives Calculator?

We can find the derivative of any inverse trig function using this calculator. Click here to get it.

How to Prove Inverse Trig Derivatives?

To prove any inverse trig derivative, we use the chain rule. For example, to find the derivative of cos -1 x, we assume that y = cos -1 x from which we get cos y = x. Differentiating both sides with respect to x, we get -sin y dy/dx = 1. From this, dy/dx = -1/sin y = -1/√ 1-cos²y = -1/√ 1-x² .

Why is the Absolute Value Sign in Inverse Trig Derivatives?

We have the absolute value sign in the derivatives of csc -1 x and sec -1 x. This is because the graph of csc -1 x is always decreasing and the graph of csc is always increasing in their domains and hence their derivatives must be always negative and positive respectively. So the absolute value sign is kept around x to keep their derivatives irrespective of x.

derivatives of inverse trig functions worksheet with answers

  • Derivatives

How to Differentiate with Inverse Trig Functions: Practice Problems

derivatives of inverse trig functions worksheet with answers

more interesting facts

derivatives of inverse trig functions worksheet with answers

more about imaginary numbers

derivatives of inverse trig functions worksheet with answers

Suppose $$f(x) = \arcsin 2x$$. Find $$f'(1/4)$$ .

Use the formula for the derivative of the inverse sine.

$$f'(x) = \frac 1 {\sqrt{1 - (2x)^2}}\cdot \frac d {dx}(2x) = \frac 2 {\sqrt{1 - 4x^2}}$$

Evaluate $$f'(1/4)$$

$$ \begin{align*} f'(1/4) & = \frac 2 {\sqrt{1 - 4\left(\frac 1 4\right)^2}}\\[6pt] & = \frac 2 {\sqrt{1 - 4\left(\frac 1 {16}\right)}}\\[6pt] & = \frac 2 {\sqrt{1 - \frac 1 4}}\\[6pt] & = \frac 2 {\sqrt{\frac 3 4}}\\[6pt] & = \frac 2 {\frac{\sqrt 3}{\sqrt 4}}\\[6pt] & = \frac 2 1 \cdot \frac{\sqrt 4}{\sqrt 3}\\[6pt] & = \frac 2 1 \cdot \frac 2 {\sqrt 3}\\[6pt] & = \frac 4 {\sqrt 3}\\[6pt] & = \frac{4\sqrt 3} 3 \end{align*} $$

$$\displaystyle f'(1/4) = \frac{4\sqrt 3} 3$$

Suppose $$f(x) = \arccos(x^2-3)$$. Find $$f'(x)$$ .

Use the formula for the derivative of the inverse cosine with the chain rule.

$$ \begin{align*} f'(x) & = \frac{-1}{\sqrt{1 - (x^2 - 3)^2}}\cdot \frac d {dx}\left(x^2 - 3\right)\\[6pt] & = \frac{-1}{\sqrt{1 - (x^2 - 3)^2}}\cdot \left(2x\right)\\[6pt] & = \frac{-2x}{\sqrt{1 - (x^2 - 3)^2}} \end{align*} $$

$$\displaystyle f'(x) = \frac{-2x}{\sqrt{1 - (x^2 - 3)^2}}$$

Suppose $$f(x) = \cot^{-1} 15x$$. Find $$f'(x)$$ .

Use the formula for the derivative of the inverse cotangent along with the chain rule.

$$ \begin{align*} f'(x) & = \frac{-1}{1+(15x)^2}\cdot \frac d {dx}(15x)\\[6pt] & = \frac{-1}{1+225x^2}\cdot (15)\\[6pt] & = \frac{-15}{1+225x^2} \end{align*} $$

$$ \displaystyle f'(x) = -\frac{15}{1+225x^2} $$

Suppose $$f(x) = \sec^{-1} \frac 3 x$$. Find $$f'(x)$$ .

Use the formula for the derivative of the inverse secant along with the chain rule.

$$ \begin{align*} f'(x) & = \frac 1 {\left|\frac 3 x\right|\sqrt{\left(\frac 3 x\right)^2 - 1}}\cdot \frac d {dx}\left(\frac 3 x\right)\\[6pt] & = \frac 1 {\frac 3{|x|}\sqrt{\frac 9 {x^2} - 1}}\cdot \frac d {dx}\left(3x^{-1}\right)\\[6pt] & = \frac 1 {\frac 3{|x|}\sqrt{\frac 9 {x^2} - 1}}\cdot \left(-3x^{-2}\right) \end{align*} $$

$$ \begin{align*} f'(x) & = \frac 1 {\frac 3{|x|}\sqrt{\frac 9 {x^2} - 1}}\cdot \left(-3x^{-2}\right)\\[6pt] & = \frac 1 {\frac 3{|x|}\sqrt{\frac 9 {x^2} - 1}}\cdot \left(\frac{-3}{x^2}\right)\\[6pt] & = \frac{-3}{\frac 3{|x|}\cdot x^2\sqrt{\frac 9 {x^2} - 1}}\\[6pt] & = \frac{-3}{3|x|\sqrt{\frac 9 {x^2} - 1}}\\[6pt] & = \frac{-1}{|x|\sqrt{\frac 1 {x^2}(9 - x^2)}}\\[6pt] & = \frac{-1}{|x|\sqrt{\frac 1 {x^2}}\cdot\sqrt{9 - x^2}}\\[6pt] & = \frac{-1}{|x|\cdot\frac 1 {\sqrt{x^2}}\cdot\sqrt{9 - x^2}}\\[6pt] & = \frac{-1}{|x|\cdot\frac 1 {|x|}\cdot\sqrt{9 - x^2}}\\[6pt] & = \frac{-1}{\sqrt{9 - x^2}} \end{align*} $$

$$\displaystyle f'(x) = \frac{-1}{\sqrt{9 - x^2}}$$

Suppose $$f(x) = x^{1/2}\arctan 9x$$. Find $$f'(1/9)$$ .

Identify the factors that make up the function.

$$ f(x) = \blue{x^{1/2}}\,\red{\arctan 9x} $$

Differentiate using the product rule.

$$ \begin{align*} f'(x) & = \blue{\frac 1 2 x^{-1/2}}\arctan 9x + x^{1/2}\cdot \red{\frac 1 {1 + (9x)^2}\cdot 9} \end{align*} $$

$$ \begin{align*} f'(x) & = \frac 1 2 \blue{x^{-1/2}}\arctan 9x + \blue{x^{1/2}}\cdot \frac 9 {1 + 81x^2}\\[6pt] & = \blue{x^{-1/2}}\left(\red{\frac 1 2} \arctan 9x + x\cdot \frac{\red 9}{1 + 81x^2}\right)\\[6pt] & = \red{\frac 1 2}x^{-1/2}\left(\arctan 9x + \frac{18x}{1 + 81x^2}\right)\\[6pt] & = \frac 1 {2x^{1/2}}\left(\arctan 9x + \frac{18x}{1 + 81x^2}\right)\\[6pt] & = \frac 1 {2\sqrt x}\left(\arctan 9x + \frac{18x}{1 + 81x^2}\right) \end{align*} $$

Evaluate $$f'(1/9)$$ .

$$ \begin{align*} f'\left(\blue{\frac 1 9}\right) & = \frac 1 {2\sqrt{\blue{\frac 1 9}}}\left(\arctan\left(9\cdot \blue{\frac 1 9}\right) + \frac{18\cdot \blue{\frac 1 9}}{1 + 81\left(\blue{\frac 1 9}\right)^2}\right)\\[6pt] & = \frac 1 {2\cdot \frac 1 3}\left(\arctan(1) + \frac 2 {1 + 81\left(\frac 1 {81}\right)}\right)\\[6pt] & = \frac 1 {2/3}\left(\frac \pi 4+ \frac 2 {1 + 1}\right)\\[6pt] & = \frac 3 2\left(\frac \pi 4 + 1\right)\\[6pt] & = \frac{3\pi} 8 + \frac 3 2\\[6pt] & \approx 2.678 \end{align*} $$

$$\displaystyle f'(1/9) = \frac{3\pi} 8 + \frac 3 2 \approx 2.678$$

Suppose $$f(x) = x^3\arccsc 4x$$. Find $$f'(x)$$

$$ f(x) = \blue{x^3}\,\red{\arccsc 4x} $$

$$ \begin{align*} f'(x) & = \blue{3x^2}\arccsc 4x + x^3\cdot\red{\frac{-4}{|4x|\sqrt{(4x)^2-1}}} \end{align*} $$

$$ \begin{align*} f'(x) & = 3x^2\arccsc 4x + x^3\cdot\frac{-4}{|4x|\sqrt{(4x)^2-1}}\\[6pt] & = 3x^2\arccsc 4x - \frac{4x^3}{4|x|\sqrt{16x^2-1}}\\[6pt] & = 3x^2\arccsc 4x - \frac{x^3}{|x|\sqrt{16x^2-1}} \end{align*} $$

$$ \displaystyle f'(x) = 3x^2\arccsc 4x - \frac{x^3}{|x|\sqrt{16x^2-1}} $$

Suppose $$f(x) = (7x+3)^{12}\cos^{-1} 2x$$. Find $$f'(x)$$ .

Identify the factors the make up the function.

$$ f(x) = \blue{(7x+3)^{12}}\,\red{\cos^{-1} 2x} $$

$$ \begin{align*} f'(x) & = \blue{12(7x+3)^{11}\cdot 7}\,\cos^{-1} 2x + (7x+3)^{12}\cdot\red{\frac 1 {\sqrt{1 - (2x)^2}}\cdot 2} \end{align*} $$

$$ \begin{align*} f'(x) & = 12(7x+3)^{11}\cdot 7\,\cos^{-1} 2x + (7x+3)^{12}\cdot\frac 1 {\sqrt{1 - (2x)^2}}\cdot 2\\[6pt] & = 84\blue{(7x+3)^{11}}\,\cos^{-1} 2x + \blue{(7x+3)^{12}}\cdot\frac 2 {\sqrt{1 - 4x^2}}\\[6pt] & = \blue{(7x+3)^{11}}\left(\red{84}\,\cos^{-1} 2x + (7x+3)\cdot\frac{\red 2}{\sqrt{1 - 4x^2}}\right)\\[6pt] & = \red 2 (7x+3)^{11}\left(42\,\cos^{-1} 2x + (7x+3)\cdot\frac 1 {\sqrt{1 - 4x^2}}\right)\\[6pt] & = 2 (7x+3)^{11}\left(42\,\cos^{-1} 2x + \frac{7x+3}{\sqrt{1 - 4x^2}}\right) \end{align*} $$

$$\displaystyle f'(x) = 2 (7x+3)^{11}\left(42\,\cos^{-1} 2x + \frac{7x+3}{\sqrt{1 - 4x^2}}\right)$$

Suppose $$f(x) = \sin 3x\,\tan^{-1} 6x$$. Find $$f'(x)$$ .

$$ f(x) = \blue{\sin 3x}\,\red{\tan^{-1} 6x} $$

$$ \begin{align*} f'(x) & = \blue{3\cos 3x}\,\tan^{-1} 6x + \sin 3x\cdot\red{\frac 1 {1 + (6x)^2}\cdot 6} \end{align*} $$

$$ \begin{align*} f'(x) & = 3\cos 3x\,\tan^{-1} 6x + \frac{6\sin 3x}{1 + 36x^2} \end{align*} $$

$$\displaystyle f'(x) = 3\cos 3x\,\tan^{-1} 6x + \frac{6\sin 3x}{1 + 36x^2}$$

Suppose $$f(x) = \arccot\left(\frac{x+4}{x-5}\right)$$. Find $$f'(2)$$ .

Differentiate with the chain rule.

$$ \begin{align*} f'(x) & = \frac{-1}{1 + \left(\frac{x+4}{x-5}\right)^2}\cdot \frac d {dx}\left(\frac{x+4}{x-5}\right) \end{align*} $$

Differentiate the quotient with the quotient rule. The parts in $$\blue{blue}$$ are related to the numerator.

$$ \begin{align*} f'(x) & = \frac{-1}{1 + \left(\frac{x+4}{x-5}\right)^2}\cdot \frac{(x-5)\cdot \blue 1 - \blue{(x+4)}\cdot 1}{(x-5)^2} \end{align*} $$

$$ \begin{align*} f'(x) & = \frac{-1}{1 + \left(\frac{x+4}{x-5}\right)^2}\cdot \frac{(x-5) - (x+4)}{(x-5)^2}\\[6pt] & = \frac{-1}{1 + \frac{(x+4)^2}{(x-5)^2}}\cdot \frac{x-5 - x-4}{(x-5)^2}\\[6pt] & = \frac{-1}{1 + \frac{(x+4)^2}{(x-5)^2}}\cdot \frac{-9}{(x-5)^2}\\[6pt] & = \frac{9}{(x-5)^2 + (x+4)^2}\\[6pt] \end{align*} $$

Evaluate $$f'(2)$$ .

$$ \begin{align*} f'(\blue 2) & = \frac{9}{(\blue 2-5)^2 + (\blue 2+4)^2}\\[6pt] & = \frac{9}{(-3)^2 + 6^2}\\[6pt] & = \frac{9}{9 + 36}\\[6pt] & = \frac 9 {45}\\[6pt] & = \frac 1 5 \end{align*} $$

$$ f'(2) = \frac 1 5 $$

Suppose $$\displaystyle f(x) = \frac{2x}{\arcsec 4x}$$. Find $$f'(x)$$ .

Differentiate using the quotient rule. The parts in $$\blue{blue}$$ are related to the numerator.

$$ \begin{align*} f'(x) & = \frac{(\arcsec 4x)\cdot \blue 2 - \blue{2x}\cdot \frac 1 {|4x|\sqrt{(4x)^2-1}}\cdot 4}{(\arcsec 4x)^2} \end{align*} $$

$$ \begin{align*} f'(x) & = \frac{(\arcsec 4x)\cdot 2 - 2x \cdot \frac 1 {|4x|\sqrt{(4x)^2-1}}\cdot 4}{(\arcsec 4x)^2}\\[6pt] & = \frac{2\arcsec 4x - \frac{8x}{4|x|\sqrt{16x^2-1}}}{(\arcsec 4x)^2}\\[6pt] & = \frac{2\arcsec 4x - \frac{2x}{|x|\sqrt{16x^2-1}}}{(\arcsec 4x)^2}\cdot \blue{\frac{|x|\sqrt{16x^2-1}}{|x|\sqrt{16x^2-1}}}\\[6pt] & = \frac{2|x|\arcsec 4x\cdot\sqrt{16x^2-1} - 2x}{|x|(\arcsec 4x)^2\sqrt{16x^2-1}} \end{align*} $$

$$ \displaystyle f'(x) = \frac{2|x|\arcsec 4x\cdot\sqrt{16x^2-1} - 2x}{|x|(\arcsec 4x)^2\sqrt{16x^2-1}} $$

Ultimate Math Solver (Free)

Free Algebra Solver ... type anything in there!

Popular pages @ mathwarehouse.com and around the web

derivatives of inverse trig functions worksheet with answers

HIGH SCHOOL

  • ACT Tutoring
  • SAT Tutoring
  • PSAT Tutoring
  • ASPIRE Tutoring
  • SHSAT Tutoring
  • STAAR Tutoring

GRADUATE SCHOOL

  • MCAT Tutoring
  • GRE Tutoring
  • LSAT Tutoring
  • GMAT Tutoring
  • AIMS Tutoring
  • HSPT Tutoring
  • ISAT Tutoring
  • SSAT Tutoring

Search 50+ Tests

Loading Page

math tutoring

  • Elementary Math
  • Pre-Calculus
  • Trigonometry

science tutoring

Foreign languages.

  • Mandarin Chinese

elementary tutoring

  • Computer Science

Search 350+ Subjects

  • Video Overview
  • Tutor Selection Process
  • Online Tutoring
  • Mobile Tutoring
  • Instant Tutoring
  • How We Operate
  • Our Guarantee
  • Impact of Tutoring
  • Reviews & Testimonials
  • Media Coverage
  • About Varsity Tutors

Calculus AB : Differentiate Inverse Trig Functions

Study concepts, example questions & explanations for calculus ab, all calculus ab resources, example questions, example question #1 : differentiate inverse trig functions.

derivatives of inverse trig functions worksheet with answers

Example Question #2 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

First, take the derivative of the function

derivatives of inverse trig functions worksheet with answers

Example Question #3 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

Example Question #4 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

This problem requires chain rule, because there is an outer and inner function.

derivatives of inverse trig functions worksheet with answers

Example Question #5 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

Example Question #6 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

To simplify further, the 10 in the numerator cancels out the 10 in the denominator. 

derivatives of inverse trig functions worksheet with answers

Example Question #7 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

Example Question #8 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

This question invokes the use of chain rule.

Example Question #9 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

Example Question #10 : Differentiate Inverse Trig Functions

derivatives of inverse trig functions worksheet with answers

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

Learning Tools by Varsity Tutors

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Mathematics LibreTexts

6.1e: Exercises - Inverse Trigonometric Functions

  • Last updated
  • Save as PDF
  • Page ID 72369

A: Concepts.

Exercise \(\PageIndex{A}\)

  • Why do the functions \(f(x)=\sin^{-1}x\) and \(g(x)=\cos^{-1}x\) have different ranges?
  • Since the functions \(y=\cos x\) and \(y=\cos^{-1}x\) are inverse functions, why is \(\cos^{-1}\left (\cos \left (-\dfrac{\pi }{6} \right ) \right )\)   not equal to \(-\dfrac{\pi }{6}\)?
  • Explain the meaning of \(\dfrac{\pi }{6}=\arcsin (0.5)\).
  • Most calculators do not have a key to evaluate \(\sec ^{-1}(2)\) .   Explain how this can be done using the cosine function or the inverse cosine function.
  • Why must the domain of the sine function, \(\sin x\) ,   be restricted to \(\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]\) for the inverse sine function to exist?
  • Discuss why this statement is incorrect: \(\arccos(\cos x)=x\) for all \(x\).
  • Determine whether the following statement is true or false and explain your answer: \(\arccos(-x)=\pi - \arccos x\)

1. The function \(y=\sin x\) is one-to-one on \(\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]\) ;   thus, this interval is the range of the inverse function of \(y=\sin x\), \(f(x)=\sin^{-1}x\)   The function \(y=\cos x\) is one-to-one on \([0,\pi ]\) ;   thus, this interval is the range of the inverse function of \(y=\cos x\), \(f(x)=\cos^{-1}x\)

3. \(\dfrac{\pi }{6}\) is the radian measure of an angle between \(-\dfrac{\pi }{2}\) and \(\dfrac{\pi }{2}\) whose sine is \(0.5\).

5. In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval \(\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]\) so that it is one-to-one and possesses an inverse.

7. True . The angle, \(\theta _1\) that equals \(\arccos(-x)\), \(x>0\), will be a second quadrant angle with reference angle, \(\theta _2\), where \(\theta _2\) equals \(\arccos x\), \(x>0\). Since \(\theta _2\) is the reference angle for \(\theta _1\), \(\theta _2=\pi - \theta _1\) and \(\arccos(-x)=\pi - \arccos x-\)add texts here. Do not delete this text first.

B: Evaluate Inverse Trigonometric Functions for "Special Angles"

Exercise \(\PageIndex{B}\)

\( \bigstar \)  Evaluate the expressions.

9.  \( 0\)     11.  \( 0\)     13.  \( 0\)     15.  \(\dfrac{\pi }{6}\)     17.  \(\dfrac{\pi }{6}\)     19.  \(\dfrac{\pi }{4}\)     21.  \(\dfrac{\pi }{6}\)      23.  \(-\dfrac{\pi }{2}\)     25.  \(-\dfrac{\pi }{4}\)    27.  \(\dfrac{2\pi }{3}\)     29.  \(\dfrac{3\pi }{4}\)     31.  \( - \dfrac{\pi }{3}\)    33.  \(-\dfrac{\pi }{3}\)     35.  \( \tan^{-1} \left(\dfrac{-1}{\sqrt{3}}\right) = - \dfrac{\pi}{6}\)  

C: Evaluate Inverse Trigonometric Functions with a Calculator

Exercise \(\PageIndex{C}\)

\( \bigstar \)  Use a calculator to evaluate each expression. Express answers to the nearest hundredth.

37.  \(1.98\)      39.  \(1.41\)      41.   \(0.73\)

D: Evaluate \( f^{-1} (f( \theta )) \) Compositions

Exercise \(\PageIndex{D}\)

\( \bigstar \)  Evaluate without a calculator.

45.  \(  \dfrac{\pi}{10}   \)       47.  \( - \dfrac{\pi}{4}   \)       49.  \(  \dfrac{\pi}{2}  \)       51.  \(  \dfrac{\pi}{10}   \)       53.  \(  \dfrac{2\pi}{3}  \)       55.  \(  \dfrac{4\pi}{5}   \)       57.  \( - \dfrac{\pi}{3}   \)       59.  \( - \dfrac{\pi}{9}   \)       61.  \( - \dfrac{\pi}{8}   \)

E: Evaluate \( f (f^{-1}( \frac{a}{b} )) \) Compositions

Exercise \(\PageIndex{E}\)

65.   \(\dfrac{1}{5}\)  \(\qquad\)  67.   \( 4 \)  \(\qquad\)  69.   undefined  

F: Evaluate \( f (g^{-1}( \frac{a}{b} )) \) Compositions

Exercise \(\PageIndex{F}\)

71.  \(\dfrac{\sqrt{5}}{5}\)            73.  \(\dfrac{4}{5}\)       75.  \(- \dfrac{2}{3}\)      77.  \(\dfrac{\sqrt{3}}{2}\)         79.  \(\dfrac{5}{13}\)  81.  \( - 2 \)           83.  \( - \dfrac{\sqrt{3}}{3}\)            85.  \( \dfrac{\sqrt{3}}{3}\)            87.  \(\dfrac{4\sqrt{7}}{7}\)       89.  \(\dfrac{3\sqrt{7}}{7}\)   

G: Evaluate \( f^{-1}(g( \theta )) \) Compositions

Exercise \(\PageIndex{G}\)

\( \bigstar \)  Find the angle \(\theta\) in the given right triangle. Round answers to the nearest hundredth.

\( \bigstar \)  Find the exact value, if possible, without a calculator, or round to the nearest hundredth. 

103.  \(0.56\)  radians          105.  \(0\)        107. undefined            109. \(  \dfrac{3\pi}{4}  \)          111.  \(0\)         113.  \(0.71\)      115.  \(-\dfrac{\pi}{4}\)       117. \(  0.86   \)  

H: Evaluate  \( f (g^{-1}( h(u) )) \) Compositions

Exercise \(\PageIndex{H}\)

For the exercises below, (a) Find the exact value of the expression in terms of \(u\). (b) State any restrictions to \(u\).

121. \( \sqrt{1-u^2} \), \( -1 \le u \le 1 \)        123. \( \dfrac{u}{\sqrt{1+u^2} } \), no restrictions        125. \( \dfrac{\sqrt{1-u^2}}{u} \), \( -1 \le u \le 1 \)      127. \(\dfrac{u-1}{\sqrt{-u^2+2u}}\), \( 0 \le u \le 2 \)        129. \(\dfrac{\sqrt{u^2-1}}{u}\), \( u \ge 1 \) or \( u \le -1 \)        131. \(\dfrac{u+\tfrac{1}{2}}{\sqrt{u^2+u+\tfrac{5}{4}}}\), no restrictions  133. \(\dfrac{u}{u+1}\), \( u \gt -\dfrac{1}{2} \)      

I: Graphs of inverses

Exercise \(\PageIndex{I}\)

138. Graph \(y=\sin^{-1} x\) and state the domain and range of the function.

139. Graph \(y=\arccos x\)   and state the domain and range of the function.

140. Graph one cycle of \(y=\tan^{-1} x\) and state the domain and range of the function.

Ex 6.3.49.png

J: Applications

Exercise \(\PageIndex{J}\)

143. Suppose a \(13\)-foot ladder is leaning against a building, reaching to the bottom of a second-floor window \(12\) feet above the ground. What angle, in radians, does the ladder make with the building?

144. Suppose you drive \(0.6\) miles on a road so that the vertical distance changes from \(0\) to \(150\) feet. What is the angle of elevation of the road?

145. An isosceles triangle has two congruent sides of length \(9\) inches. The remaining side has a length of \(8\) inches. Find the angle that a side of \(9\) inches makes with the \(8\)-inch side.

146. Without using a calculator, approximate the value of \(\arctan (10,000)\) .   Explain why your answer is reasonable.

147. A truss for the roof of a house is constructed from two identical right triangles. Each has a base of \(12\) feet and height of \(4\) feet. Find the measure of the acute angle adjacent to the \(4\)-foot side.

148. The line \(y=\dfrac{3}{5}x\) passes through the origin in the \(x,y\)-plane. What is the measure of the angle that the line makes with the positive \(x\)-axis?

149. The line \(y=\dfrac{-3}{7}x\) passes through the origin in the \(x,y\) -plane. What is the measure of the angle that the line makes with the negative \(x\)-axis?

150. What percentage grade should a road have if the angle of elevation of the road is \(4\) degrees? (The percentage grade is defined as the change in the altitude of the road over a \(100\)-foot horizontal distance. For example a \(5\%\) grade means that the road rises \(5\) feet for every \(100\) feet of horizontal distance.)

151. A \(20\)-foot ladder leans up against the side of a building so that the foot of the ladder is \(10\) feet from the base of the building. If specifications call for the ladder's angle of elevation to be between \(35\) and \(45\) degrees, does the placement of this ladder satisfy safety specifications?

152. Suppose a \(15\)-foot ladder leans against the side of a house so that the angle of elevation of the ladder is \(42\) degrees. How far is the foot of the ladder from the side of the house?

143.  \(0.395\)  radians         145.  \(1.11\)  radians      147.  \(1.25\)  radians      149.  \(0.405\)  radians    151.  No. The angle the ladder makes with the horizontal is \(60\) degrees..

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

AP®︎/College Calculus AB

Course: ap®︎/college calculus ab   >   unit 3, derivatives of inverse functions.

  • Derivatives of inverse functions: from equation
  • Derivatives of inverse functions: from table
  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
  • an exact decimal, like 0.75 ‍  
  • a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
  • Math Article
  • Derivative Inverse Trigonometric Functions

Derivative of Inverse Trigonometric functions

The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. These functions are used to obtain angle for a given trigonometric value. Inverse trigonometric functions have various application in engineering, geometry, navigation etc.

Representation of functions:

Generally, the inverse trigonometric function are represented by adding arc in prefix for a trigonometric function, or by adding the power of -1, such as:

Inverse of sin x = arcsin(x) or \(\begin{array}{l}\sin^{-1}x\end{array} \)

Let us now find the derivative of Inverse trigonometric function

Example: Find the derivative of a function \(\begin{array}{l}y = \sin^{-1}x\end{array} \) .

Solution:Given \(\begin{array}{l}y = \sin^{-1}x\end{array} \) …………(i)

\(\begin{array}{l}\Rightarrow x = \sin y\end{array} \)

Differentiating the above equation w.r.t. x, we have:

\(\begin{array}{l}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\cos y}\end{array} \)

Putting the value of y form (i), we get

\(\begin{array}{l}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\cos y} = \frac{1}{\cos (\sin^{-1}x)}\end{array} \) ………..(ii)

From equation (ii), we can see that the value of cos y cannot be equal to 0, as the function would become undefined.

\(\begin{array}{l}\Rightarrow \sin^{-1}x \neq \frac{-\pi}{2}, \frac{\pi}{2}\end{array} \)

i.e. \(\begin{array}{l}x \neq -1,1\end{array} \)

From (i) we have \(\begin{array}{l}y = \sin^{-1}x\end{array} \)

\(\begin{array}{l}\Rightarrow \sin y = \sin (\sin^{-1}x)\end{array} \)

Using property of trigonometric function,

\(\begin{array}{l}\cos^{2}y = 1 – \sin^{2}y = 1 – (\sin (\sin^{-1}x))^{2} = 1 – x^{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \cos y = \sqrt{1 – x^{2}}\end{array} \) …………(iii)

Now putting the value of (iii) in (ii), we have

\(\begin{array}{l}\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\sqrt{1-x^{2}}}\end{array} \)

Therefore, the Derivative of Inverse sine function is

\(\begin{array}{l}\frac{\mathrm{d} }{\mathrm{d} x} (\sin^{-1}x)= \frac{1}{\sqrt{1-x^{2}}}\end{array} \)

Derivatives of Inverse trigonometric functions

Video lesson on trigonometry.

derivatives of inverse trig functions worksheet with answers

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Maths related queries and study materials

Your result is as below

Request OTP on Voice Call

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Post My Comment

derivatives of inverse trig functions worksheet with answers

Nice explain 😃 thank you so much for all byjus Teachers 🙏🖤

derivatives of inverse trig functions worksheet with answers

  • Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

close

IMAGES

  1. Derivatives of Inverse Trig Functions (2 WS, 20 problems with solutions)

    derivatives of inverse trig functions worksheet with answers

  2. Inverse Trig Derivatives (Derivatives of Inverse Trig Functions)

    derivatives of inverse trig functions worksheet with answers

  3. Inverse Trig Derivatives

    derivatives of inverse trig functions worksheet with answers

  4. Derivatives Of Inverse Functions Worksheet With Answers Pdf

    derivatives of inverse trig functions worksheet with answers

  5. Quiz & Worksheet

    derivatives of inverse trig functions worksheet with answers

  6. Derivatives Of Inverse Trig Functions Worksheet

    derivatives of inverse trig functions worksheet with answers

VIDEO

  1. Inverse Trig Derivatives

  2. The Derivatives Of Inverse Trig Functions

  3. Derivatives of Inverse Trig Functions

  4. Trig Derivatives #mathemagics

  5. Derivatives of Inverse Trig Functions 1

  6. Derivatives of Inverse Trig Functions

COMMENTS

  1. Calculus I

    Section 3.7 : Derivatives of Inverse Trig Functions For each of the following problems differentiate the given function. T (z) = 2cos(z) +6cos−1(z) T ( z) = 2 cos ( z) + 6 cos − 1 ( z) Solution g(t) = csc−1(t) −4cot−1(t) g ( t) = csc − 1 ( t) − 4 cot − 1 ( t) Solution y = 5x6−sec−1(x) y = 5 x 6 − sec − 1 ( x) Solution

  2. PDF Differentiation

    Differentiation - Inverse Trigonometric Functions. Differentiate each function with respect to x. 1) y = −1 cos −5 x3. −1 3) y = tan 2 x4. −1 5) y = (sin 5 x2)3. −1 7) y = (cos 4 x2)2. ©9 L2d0J1s2x hKLuAtaaf qSHoIf3t2wKaBr4eH nLZLzCs.N k qAilul5 NroiYghhZtDsN WrzezsRecr9vverdF.r C 2MEatdseN Ww4i2tuhc VIenIfeiBnMiVtaeU ...

  3. PDF CHAPTER 25 Derivatives of Inverse Trig Functions

    We will derive six new derivative formulas for the six inverse trigonometric functions: dxhsin°1(x)i d dxhtan°1(x)i d dxhsec°1(x)i d dxhcos°1(x)i d dxhcot°1(x)i d dxhcsc°1(x)i d These formulas will flow from the inverse rule from Chapter 24 (page 278): °1(x)i 1 = °f °1(x)¢. 0 25.1 Derivatives of Inverse Sine and Cosine

  4. PDF Derivatives of inverse function PROBLEMS and SOLUTIONS

    Derivative of the inverse function at a point is the reciprocal of the derivative of the function at the corresponding point. Slope of the line tangent to at = is the reciprocal of the slope of at = . -1 1. Find tangent line at point (4, 2) of the graph of f if f(x) = x3 + 2x - 8 2.

  5. PDF Mr. Yang's Teacher Website

    Calculus Ch. 5.6 Inverse Trig Derivatives Classwork Worksheet THEOREM 5.16 Derivatives of Inverse Trigonometric Functions Let u be a differentiable function of x. arcsin u [arctan u] — [arcsec u] ul 112 arccos u [arccot u] [arccsc u] Evaluating an Expression In Exercises 21-24, evaluate each expression without using a calculator.

  6. PDF AP CALCULUS AB/BC: Inverse Trig Derivatives| WORKSHEET

    Answer Key to Worksheet: Inverse Trig Derivatives Note: P12 and P14 are equations of tangent lines. The rest are derivatives. Please report any mistakes you find. VL-CLSU X] cac[x] 24U[X] x X] x T x] 5 X] VL-C241J[X] X . Title: AP CALCULUS AB/BC: Inverse Trig Derivatives| WORKSHEET

  7. PDF 03

    Derivatives of Inverse Functions Date________________ Period____ For each problem, find ( f −1) ' ( x) by direct computation. 1) f ( x) = −3 x + 3 2) f ( x) = −2 x + 3 For each problem, find ( f −1) 1 ' ( x) by using the theorem ( f −1) ' ( x) =

  8. Calculus I

    In order to derive the derivatives of inverse trig functions we'll need the formula from the last section relating the derivatives of inverse functions. If f(x) and g(x) are inverse functions then, g ′ (x) = 1 f ′ (g(x)) Recall as well that two functions are inverses if f(g(x)) = x and g(f(x)) = x.

  9. Derivatives of inverse trigonometric functions

    Derivatives of inverse trigonometric functions Google Classroom You might need: Calculator h ( x) = arctan ( − x 2) h ′ ( − 7) = Use an exact expression. Show Calculator Stuck? Review related articles/videos or use a hint. Report a problem Do 4 problems

  10. Inverse Trig Derivatives (Derivatives of Inverse Trig Functions)

    The inverse trig derivatives are the derivatives of the inverse trigonometric functions arcsin (or sin -1 ), arccos (or cos -1 ), arctan (or tan -1 ), etc. We use implicit differentiation to find the derivatives of the inverse trig function which we we explore in detail in the upcoming section. Here are the inverse trig derivatives:

  11. PDF Differentiation

    Differentiation - Trigonometric Functions Date________________ Period____ Differentiate each function with respect to x. 1) f ( x) = sin 2 x3 3) y = sec 4 x5 5) y = ( 2 x5 + 3)cos x2 2) y = tan 5 x3 4) y = csc 5 x5 −2 x2 − 5 6) y = cos 2 x3 7) f ( x) 3 = sin x5 8) f ( x) = cos ( −3 x2 + 2)2

  12. PDF Inverse Trig Functions

    -2- Worksheet by Kuta Software LLC Find the exact value of each expression. 9) csc (cos ) 10) cos sec 11) csc sec 12) sin (sec ) 13) sec cot 14) tan (sec ) Identify the domain and range of each. Then sketch the graph. 15) y sin cot x x y

  13. PDF 3.4 Differentiating Inverse Trigonometric Functions

    3.4 - Differentiating Inverse Trigonometric Functions Find the derivative of each function. Oh no! A murder has been committed. Peppa Pig has been found dead and the police have no leads. It is up to you solve this mystery. Solve each of the following multiple-choice questions by taking the derivative of each inverse trigonometric function.

  14. Derivatives of the Inverse Trigonometric Functions

    so. dy dx = 1 cosy = 1 √1 − x2. Thus we have found the derivative of y = arcsinx, d dx (arcsinx) = 1 √1 − x2. Exercise 1. Use the same approach to determine the derivatives of y = arccosx, y = arctanx, and y = arccotx. Answer. Example 2: Finding the derivative of y = arcsecx. Find the derivative of y = arcsecx.

  15. How to Differentiate with Inverse Trig Functions

    Working with derivatives of inverse trig functions. 10 interactive practice Problems worked out step by step. Chart Maker; Games; Math Worksheets; Learn to code with Penjee; Toggle navigation. Gifs; Algebra; Geometry; Trig; Calc; Teacher Tools; Learn to Code ... Show Answer. Step 1. Use the formula for the derivative of the inverse sine. Step 1 ...

  16. 3.4 Differentiating Inverse Trigonometric Functions

    3.4 Differentiating Inverse Trigonometric Functions: Next Lesson. Packet. calc_3.4_packet.pdf: File Size: 446 kb: File Type: pdf: Download File. Want to save money on printing? Support us and buy the Calculus workbook with all the packets in one nice spiral bound book. Solution manuals are also available. Practice Solutions.

  17. Differentiate Inverse Trig Functions

    Correct answer: − 4 65. Explanation: f(x) = cot−1(4x) First, take the derivative of the function. f′(x) = − 4 1 + (4x)2 = − 4 1 + 16x2. Especially when given inverse trigonometry derivative questions, be on the lookout for multiple functions embedded in the same problem. For example, in this problem there is both an outer function ...

  18. PDF Derivatives and Integrals of Inverse Trig Functions

    -2-Worksheet by Kuta Software LLC Evaluate each indefinite integral. 7) ∫ 1 4 + x2 dx 8) 1 xx2 − 9 dx 9) ∫ 1 xx2 − 25 dx 10) ∫ 1 9 − x2 dx 11) ∫ 2e 2x 25 − e4x dx 12) ∫ 9x 25 + 9x6 dx 13) ∫ 3e3x e3xe6x − 1 dx 14) 12x2 4x316x6 − 9 dx

  19. 6.1e: Exercises

    Exercise 6.1e. A. Why do the functions f(x) = sin − 1x and g(x) = cos − 1x have different ranges? Since the functions y = cosx and y = cos − 1x are inverse functions, why is cos − 1(cos( − π 6)) not equal to − π 6? Explain the meaning of π 6 = arcsin(0.5). Most calculators do not have a key to evaluate sec − 1(2) Explain how ...

  20. Derivatives of inverse functions (practice)

    Derivatives of inverse functions. Let g and h be inverse functions. The following table lists a few values of g , h , and h ′ . Stuck? Review related articles/videos or use a hint. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.

  21. Calculus I

    Solution. Where in the range [−2,7] [ − 2, 7] is the function f (x) =4cos(x) −x f ( x) = 4 cos. ⁡. ( x) − x is increasing and decreasing. Solution. Here is a set of practice problems to accompany the Derivatives of Trig Functions section of the Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University.

  22. PDF Calculus 2 Tutor Worksheet 2 Derivatives of Inverse Trigonometric Functions

    3. Domains of derivatives of inverse trigonometric functions. 3(a). Answer: 1 <x<1 . The domain of f(x) = cos 1 xis 1 x 1. It is the same as the domain of f(x) = sin 1 xthat we found previously. The domain of df dx = 1 p 1 x2 is all xwhere it is defined; that is, where 1 x2 0 and p 1 2x 6= 0

  23. Derivative of Inverse Trigonometric functions

    Generally, the inverse trigonometric function are represented by adding arc in prefix for a trigonometric function, or by adding the power of -1, such as: Inverse of sin x = arcsin (x) or. \ (\begin {array} {l}\sin^ {-1}x\end {array} \) Let us now find the derivative of Inverse trigonometric function. Example: Find the derivative of a function.