What you’ll learn

How to differentiate inverse trig functions using the chain rule or by simplifying

Inverse trigonometric functions

Regular trigonometric functions take an angle $θ$ as the input and output a trigonometric ratio. In a right-triangle setting, that ratio is “opposite over hypotenuse” for sine, “adjacent over hypotenuse” for cosine, and so on (SOHCAHTOA).

Inverse trigonometric functions reverse that relationship: you input a trigonometric ratio, and the output is the angle that produces it (within the function’s restricted range). For example,

$sin (\frac{π}{4}) = \frac{2}{2} ⟺ sin^{- 1} (\frac{2}{2}) = \frac{π}{4}$

To find the derivative of the inverse sine function ( $sin^{- 1} (x)$ or $arcsin (x)$ ), start with the equivalent relationship

$y = sin^{- 1} (x) ⟺ x = sin (y)$

Differentiate implicitly with respect to $x$ :

$1 = cos (y) \cdot y^{'}$

$y^{'} = \frac{1}{cos ( y )}$

$= \frac{1}{cos ( sin ^{- 1} ( x ))}$

Now simplify the composite trig expression. The output of $sin^{- 1} (x)$ is an angle, so let

$sin^{- 1} (x) = θ ⟺ x = sin (θ)$

Think of $x$ as $\frac{x}{1}$ . In a right triangle with angle $θ$ :

opposite side $= x$
hypotenuse $= 1$

By the Pythagorean theorem, the adjacent side is $1 - x^{2}$ .

Using this triangle,

$y^{'} = \frac{1}{cos ( sin ^{- 1} ( x ))}$

$= \frac{1}{cos ( θ )}$

$= \frac{1}{1 - x ^{2}}$

So, for $y = sin^{- 1} (x)$ ,

$y^{'} = \frac{1}{1 - x ^{2}}$

Repeating this same idea (implicit differentiation, then simplifying with a triangle when needed) gives the derivatives of the six inverse trig functions:

$f (x)$	$f^{'} (x)$
$sin^{- 1} (x)$	$\frac{1}{1 - x ^{2}}$
$cos^{- 1} (x)$	$\frac{- 1}{1 - x ^{2}}$
$tan^{- 1} (x)$	$\frac{1}{1 + x ^{2}}$
$cot^{- 1} (x)$	$\frac{- 1}{1 + x ^{2}}$
$sec^{- 1} (x)$	$\frac{1}{∣ x ∣ x ^{2} - 1}$
$csc^{- 1} (x)$	$\frac{- 1}{∣ x ∣ x ^{2} - 1}$

Notice that every other derivative is just the negative of the one above it.

For $sec^{- 1} (x)$ , the absolute value ensures the derivative stays positive for all $x$ in the domain (consistent with the graph’s tangent-line behavior). For $csc^{- 1} (x)$ , the absolute value similarly supports the fact that the derivative stays negative on its domain.

Notation pitfall: $sin^{- 1} (x)$ means arcsine, the inverse function - not $\frac{1}{sin ( x )} = csc (x)$ . The derivative of $sin^{- 1} (x)$ is $\frac{1}{1 - x ^{2}}$ , not $- csc (x) cot (x)$ .

Examples with chain rule

If an inverse trig function has an argument that isn’t just $x$ , do two things:

Replace every $x$ in the standard derivative formula with the new argument.
Multiply by the derivative of that argument (this is the chain rule).

For example,

Example 1. Find the derivative of

$y = sec^{- 1} (x^{2} + 1)$

Solution

The derivative of the basic function $y = sec^{- 1} (x)$ is

$y^{'} = \frac{1}{∣ x ∣ x ^{2} - 1} .$

Here the argument is $x^{2} + 1$ . Substitute $x^{2} + 1$ everywhere you see $x$ , then multiply by the derivative of $x^{2} + 1$ (which is $2 x$ ):

$y^{'} = \frac{1}{∣ x ^{2} + 1∣ ( x ^{2} + 1 ) ^{2} - 1} \cdot 2 x$

Since $x^{2} + 1 > 0$ for all real $x$ , the absolute value can be dropped:

$y^{'} = \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 1 ) ^{2} - 1}$

Example 2. Find $\frac{d ^{2} y}{d x ^{2}}$ for

$y = cot^{- 1} (e^{x})$

Solution

(spoiler)

The derivative of $y = cot^{- 1} (x)$ is $\frac{- 1}{1 + x ^{2}}$ .

With argument $e^{x}$ , the first derivative is

$y^{'} = \frac{- 1}{1 + ( e ^{x} ) ^{2}} \cdot e^{x}$

$= \frac{- e ^{x}}{1 + e ^{2 x}}$

Now take the second derivative using the quotient rule:

$y^{''} = \frac{( 1 + e ^{2 x} ) ( - e ^{x} ) - ( - e ^{x} ) ( 2 e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}}$

$= \frac{( - e ^{x} ) ( 1 + e ^{2 x} - 2 e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}}$

$= \frac{- e ^{x} ( 1 - e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}}$

$= \frac{e ^{x} ( e ^{2 x} - 1 )}{( 1 + e ^{2 x} ) ^{2}}$

Example 3. Find the derivative of

$f (x) = tan (cos^{- 1} (2 x))$

Solution

(spoiler)

There are two ways to do this:

Simplify the nested trig function first by drawing a triangle and then differentiate.
Differentiate first and then simplify using a triangle.

We’ll show both.

Approach 1: Simplify $\to$ differentiate

Let $cos^{- 1} (2 x) = θ$ .

Then

$cos (θ) = \frac{2 x}{1} .$

Using the Pythagorean theorem, the remaining side of the triangle is $1 - 4 x^{2}$ .

Then

$f (x) = tan (θ) = \frac{1 - 4 x ^{2}}{2 x} .$

Differentiate using the quotient rule:

$f^{'} (x) = \frac{2 x \cdot \frac{1}{2 1 - 4 x ^{2}} \cdot ( - 8 x ) - 1 - 4 x ^{2} \cdot 2}{( 2 x ) ^{2}}$

$= \frac{\frac{- 16 x ^{2}}{2 1 - 4 x ^{2}} - 2 1 - 4 x ^{2}}{4 x ^{2}}$

$= \frac{\frac{- 8 x ^{2}}{1 - 4 x ^{2}} - \frac{2 ( 1 - 4 x ^{2} )}{1 - 4 x ^{2}}}{4 x ^{2}}$

$= \frac{\frac{- 8 x ^{2} - 2 + 8 x ^{2}}{1 - 4 x ^{2}}}{4 x ^{2}}$

$= - \frac{2}{4 x ^{2} 1 - 4 x ^{2}}$

$= - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

Approach 2: Differentiate $\to$ simplify

Differentiate using the chain rule:

$f^{'} (x) = sec^{2} (cos^{- 1} (2 x)) \cdot \frac{- 1}{1 - ( 2 x ) ^{2}} \cdot 2$

To simplify $sec^{2} (cos^{- 1} (2 x))$ , use the same triangle as above. Since $cos (θ) = 2 x$ , we have $sec (θ) = \frac{1}{2 x}$ , so

$sec^{2} (θ) = \frac{1}{4 x ^{2}} .$

Then

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

$= - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

In many problems, the faster approach is the one that avoids the quotient rule.

Example 4. Find the equation of the tangent line to $y = arccos (x^{2})$ at $x = \frac{2}{2}$ .

Solution

(spoiler)

First find the point of tangency by evaluating $y$ at $x = \frac{2}{2}$ :

$y = arccos ((\frac{2}{2})^{2}) = arccos (\frac{1}{2}) .$

Rewrite in cosine form:

$cos^{- 1} (1/2) = y ⟺ cos (y) = 1/2.$

From the unit circle (or a $30°$ - $60°$ - $90°$ triangle), $y = \frac{π}{3}$ .

So the point of tangency is

$(\frac{2}{2}, \frac{π}{3}) .$

Next find the slope by differentiating $y = cos^{- 1} (x^{2})$ :

$\frac{d}{d x} cos^{- 1} (x^{2})$

$= - \frac{1}{1 - ( x ^{2} ) ^{2}} \cdot 2 x$

$= - \frac{2 x}{1 - x ^{4}}$

Evaluate at $x = \frac{2}{2}$ :

$- \frac{2 \cdot \frac{2}{2}}{1 - ( \frac{2}{2} ) ^{4}}$

$= - \frac{2}{1 - \frac{4}{16}}$

$= - \frac{2}{\frac{3}{4}}$

$= - \frac{2 2}{3}$

$= - \frac{2 6}{3}$

Now use point-slope form with point $(\frac{2}{2}, \frac{π}{3})$ and slope $- \frac{2 6}{3}$ :

$y - \frac{π}{3} = - \frac{2 6}{3} (x - \frac{2}{2})$

What you’ll learn

How to differentiate inverse trig functions using the chain rule or by simplifying

Inverse trigonometric functions

Inverse trigonometric functions reverse that relationship: you input a trigonometric ratio, and the output is the angle that produces it (within the function’s restricted range). For example,

$sin (\frac{π}{4}) = \frac{2}{2} ⟺ sin^{- 1} (\frac{2}{2}) = \frac{π}{4}$

To find the derivative of the inverse sine function ( $sin^{- 1} (x)$ or $arcsin (x)$ ), start with the equivalent relationship

$y = sin^{- 1} (x) ⟺ x = sin (y)$

Differentiate implicitly with respect to $x$ :

$1 = cos (y) \cdot y^{'}$

$y^{'} = \frac{1}{cos ( y )}$

$= \frac{1}{cos ( sin ^{- 1} ( x ))}$

Now simplify the composite trig expression. The output of $sin^{- 1} (x)$ is an angle, so let

$sin^{- 1} (x) = θ ⟺ x = sin (θ)$

Think of $x$ as $\frac{x}{1}$ . In a right triangle with angle $θ$ :

opposite side $= x$
hypotenuse $= 1$

By the Pythagorean theorem, the adjacent side is $1 - x^{2}$ .

Using this triangle,

$y^{'} = \frac{1}{cos ( sin ^{- 1} ( x ))}$

$= \frac{1}{cos ( θ )}$

$= \frac{1}{1 - x ^{2}}$

So, for $y = sin^{- 1} (x)$ ,

$y^{'} = \frac{1}{1 - x ^{2}}$

Repeating this same idea (implicit differentiation, then simplifying with a triangle when needed) gives the derivatives of the six inverse trig functions:

$f (x)$	$f^{'} (x)$
$sin^{- 1} (x)$	$\frac{1}{1 - x ^{2}}$
$cos^{- 1} (x)$	$\frac{- 1}{1 - x ^{2}}$
$tan^{- 1} (x)$	$\frac{1}{1 + x ^{2}}$
$cot^{- 1} (x)$	$\frac{- 1}{1 + x ^{2}}$
$sec^{- 1} (x)$	$\frac{1}{∣ x ∣ x ^{2} - 1}$
$csc^{- 1} (x)$	$\frac{- 1}{∣ x ∣ x ^{2} - 1}$

Notice that every other derivative is just the negative of the one above it.

Examples with chain rule

If an inverse trig function has an argument that isn’t just $x$ , do two things:

Replace every $x$ in the standard derivative formula with the new argument.
Multiply by the derivative of that argument (this is the chain rule).

For example,

Example 1. Find the derivative of

$y = sec^{- 1} (x^{2} + 1)$

Solution

The derivative of the basic function $y = sec^{- 1} (x)$ is

$y^{'} = \frac{1}{∣ x ∣ x ^{2} - 1} .$

Here the argument is $x^{2} + 1$ . Substitute $x^{2} + 1$ everywhere you see $x$ , then multiply by the derivative of $x^{2} + 1$ (which is $2 x$ ):

$y^{'} = \frac{1}{∣ x ^{2} + 1∣ ( x ^{2} + 1 ) ^{2} - 1} \cdot 2 x$

Since $x^{2} + 1 > 0$ for all real $x$ , the absolute value can be dropped:

$y^{'} = \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 1 ) ^{2} - 1}$

Example 2. Find $\frac{d ^{2} y}{d x ^{2}}$ for

$y = cot^{- 1} (e^{x})$

Solution

(spoiler)

The derivative of $y = cot^{- 1} (x)$ is $\frac{- 1}{1 + x ^{2}}$ .

With argument $e^{x}$ , the first derivative is

$y^{'} = \frac{- 1}{1 + ( e ^{x} ) ^{2}} \cdot e^{x}$

$= \frac{- e ^{x}}{1 + e ^{2 x}}$

Now take the second derivative using the quotient rule:

$y^{''} = \frac{( 1 + e ^{2 x} ) ( - e ^{x} ) - ( - e ^{x} ) ( 2 e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}}$

$= \frac{( - e ^{x} ) ( 1 + e ^{2 x} - 2 e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}}$

$= \frac{- e ^{x} ( 1 - e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}}$

$= \frac{e ^{x} ( e ^{2 x} - 1 )}{( 1 + e ^{2 x} ) ^{2}}$

Example 3. Find the derivative of

$f (x) = tan (cos^{- 1} (2 x))$

Solution

(spoiler)

There are two ways to do this:

Simplify the nested trig function first by drawing a triangle and then differentiate.
Differentiate first and then simplify using a triangle.

We’ll show both.

Approach 1: Simplify $\to$ differentiate

Let $cos^{- 1} (2 x) = θ$ .

Then

$cos (θ) = \frac{2 x}{1} .$

Using the Pythagorean theorem, the remaining side of the triangle is $1 - 4 x^{2}$ .

Then

$f (x) = tan (θ) = \frac{1 - 4 x ^{2}}{2 x} .$

Differentiate using the quotient rule:

$f^{'} (x) = \frac{2 x \cdot \frac{1}{2 1 - 4 x ^{2}} \cdot ( - 8 x ) - 1 - 4 x ^{2} \cdot 2}{( 2 x ) ^{2}}$

$= \frac{\frac{- 16 x ^{2}}{2 1 - 4 x ^{2}} - 2 1 - 4 x ^{2}}{4 x ^{2}}$

$= \frac{\frac{- 8 x ^{2}}{1 - 4 x ^{2}} - \frac{2 ( 1 - 4 x ^{2} )}{1 - 4 x ^{2}}}{4 x ^{2}}$

$= \frac{\frac{- 8 x ^{2} - 2 + 8 x ^{2}}{1 - 4 x ^{2}}}{4 x ^{2}}$

$= - \frac{2}{4 x ^{2} 1 - 4 x ^{2}}$

$= - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

Approach 2: Differentiate $\to$ simplify

Differentiate using the chain rule:

$f^{'} (x) = sec^{2} (cos^{- 1} (2 x)) \cdot \frac{- 1}{1 - ( 2 x ) ^{2}} \cdot 2$

To simplify $sec^{2} (cos^{- 1} (2 x))$ , use the same triangle as above. Since $cos (θ) = 2 x$ , we have $sec (θ) = \frac{1}{2 x}$ , so

$sec^{2} (θ) = \frac{1}{4 x ^{2}} .$

Then

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

$= - \frac{1}{2 x ^{2} 1 - 4 x ^{2}}$

In many problems, the faster approach is the one that avoids the quotient rule.

Example 4. Find the equation of the tangent line to $y = arccos (x^{2})$ at $x = \frac{2}{2}$ .

Solution

(spoiler)

First find the point of tangency by evaluating $y$ at $x = \frac{2}{2}$ :

$y = arccos ((\frac{2}{2})^{2}) = arccos (\frac{1}{2}) .$

Rewrite in cosine form:

$cos^{- 1} (1/2) = y ⟺ cos (y) = 1/2.$

From the unit circle (or a $30°$ - $60°$ - $90°$ triangle), $y = \frac{π}{3}$ .

So the point of tangency is

$(\frac{2}{2}, \frac{π}{3}) .$

Next find the slope by differentiating $y = cos^{- 1} (x^{2})$ :

$\frac{d}{d x} cos^{- 1} (x^{2})$

$= - \frac{1}{1 - ( x ^{2} ) ^{2}} \cdot 2 x$

$= - \frac{2 x}{1 - x ^{4}}$

Evaluate at $x = \frac{2}{2}$ :

$- \frac{2 \cdot \frac{2}{2}}{1 - ( \frac{2}{2} ) ^{4}}$

$= - \frac{2}{1 - \frac{4}{16}}$

$= - \frac{2}{\frac{3}{4}}$

$= - \frac{2 2}{3}$

$= - \frac{2 6}{3}$

Now use point-slope form with point $(\frac{2}{2}, \frac{π}{3})$ and slope $- \frac{2 6}{3}$ :

$y - \frac{π}{3} = - \frac{2 6}{3} (x - \frac{2}{2})$

Inverse trig derivatives

What you’ll learn

Inverse trigonometric functions

Examples with chain rule

Solution

Solution

Solution

Solution

Inverse trig derivatives

What you’ll learn

Inverse trigonometric functions

Examples with chain rule

Solution

Solution

Solution

Solution