Inverse trig derivatives

Inverse trigonometric functions

Regular trigonometric functions take an angle $\theta$ as the input, and output a trigonometric ratio. In the context of triangles, the ratio would be the “opposite side over the hypotenuse” for sine, “adjacent side over hypotenuse” for cosine, etc. (remember the mnemonic SOHCAHTOA).

Inverse trigonometric functions swap the input and output - given a trigonometric ratio as the input, the output will be the angle that produces it, but only in certain ranges depending on the function. For example,

$$\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}⟺{\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi }{4}$$

To find the derivative of the inverse sine $({\sin }^{-1}(x)$ or $\arcsin (x))$ function, we use the fact that

$$y={\sin }^{-1}(x)⟺x=\sin (y)$$

Implicitly differentiating $x=\sin (y)$,

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

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

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

To simplify this composite trig function, note that the output of the ${\sin }^{-1}(x)$ function is an angle (call it $\theta$).

$${\sin }^{-1}(x)=\theta ⟺x=\sin (\theta )$$

We can then draw a triangle that depicts this relationship. $x$ is the same thing as $\frac{x}{1}$, so in the triangle, the side opposite to $\theta$ is $x$, and the hypotenuse is $1$. Then using the Pythagorean theorem, the length of the remaining unknown leg is $\sqrt{1-x^2}$.

Using the triangle shown,

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

$$=\frac{1}{\cos (\theta )}$$

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

So the derivative of the $\arcsin (x)$ function, $y={\sin }^{-1}(x)$, is

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

Repeating this process of simplifying nested trig functions and drawing triangles results in the derivatives of the six inverse trig functions:

Notice that every other one is simply the negative version of the one above it.

Because the value of the derivative is positive for all $x$ on ${\sec }^{-1}(x)$ (look at the graph and how all tangent lines to it are oriented), the absolute value is just to ensure this is true for all $x$ in the domain.

The same applies to the absolute value on ${\csc }^{-1}(x)$ - for all $x$ in the domain of the inverse trig function, the derivative stays negative.

Examples with chain rule

If an inverse trig function has an argument that isn’t just plain $x$, then all instances of $x$ in the derivative expressions above should be replaced with the actual argument of the inverse trig function. Don’t forget to use the chain rule as well (multiplying by the derivative of the argument). For example,

1. Find the derivative of

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

Solution

The derivative of the basic $y={\sec }^{-1}(x)$ function is $y^{\prime }=\frac{1}{|x|\sqrt{x^2-1}}$.

For $y={\sec }^{-1}(x^2+1)$, however,

$$y^{\prime }=\boxed{\frac{1}{|x^2+1|\sqrt{(x^2+1{)}^2-1}}\cdot 2x}$$

where each instance of $x$ is replaced by the new argument $x^2+1$ and the chain rule is applied by multiplying the derivative of the inner function $x^2+1$, or $2x$.

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

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

Solution

3. Find the derivative of

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

Solution

(spoiler)

There are two ways to do this:

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

We’ll show both.

Approach 1: Simplify $\to$ differentiate

Let ${\cos }^{-1}(2x)=\theta$.

Then $\cos (\theta )=\frac{2x}{1}$. Using the Pythagorean theorem, the remaining side of the triangle is $\sqrt{1-4x^2}$, as shown.

Then $f(x)$ simplifies to $\tan (\theta )=\frac{\sqrt{1-4x^2}}{2x}$.

Differentiating this with the quotient rule,

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

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

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

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

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

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

Approach 2: Differentiate $\to$ simplify

Differentiating with the chain rule,

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

To simplify ${\sec }^2({\cos }^{-1}(2x))$, use the same triangle shown above.

$${\sec }^2(\theta )=\frac{1}{4x^2}$$

Then

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

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

In most cases, the faster approach will be the one that avoids the quotient rule.

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

Solution