Inverse trig derivatives | Advanced differentiation

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.

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,

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 $2 x$ ):

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

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:

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

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

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

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

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.

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,

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 $2 x$ ):

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

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:

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

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

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

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

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.

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

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