Introduction

Now that we know how to differentiate inverse functions, the next logical step is to find the derivatives of some common inverse functions. Specifically the inverse trigonometric functions. These derivatives can be found using the methods we learned earlier in this unit.

Finding the Derivatives of the Inverse Trig Functions

I am not going to go over the calculations for each of the inverse trigonometric functions and the inverse reciprocal trigonometric functions. But just to show how these are derived in general, I will go over an example of finding the derivative of the inverse sine of $x$. First let's let ${\sin }^{-1}(x)=y$ and $\sin (y)=x$. Then we can differentiate $\sin (y)=x$ implicitly with respect to $x$ to find the derivative of ${\sin }^{-1}(x)$ or $\arcsin (x)$:

$\sin (y)=x$

$\frac{dy}{dx}[\sin (y)]=\frac{dy}{dx}[x]$

$\cos (y)\cdot \frac{dy}{dx}=1$$\frac{dy}{dx}=\frac{1}{\cos (y)}$

We have found the derivative of $\arcsin (x)$ with respect to $x$ to be equal to $\frac{1}{\cos (y)}$. The problem with this though is that our derivative is in terms of $y$. To solve this we need to apply the pythagorean identity of sine and cosine. Earlier we defined that $\sin (y)=x$, so according to the pythagorean identity ${\sin }^2(y)+{\cos }^2(y)=1$. Since we know that $\sin (y)=x$ we can substitute in $x$ for $\sin (y)$ and then just solve for $\cos (y)$: 

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

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

$\cos (y)=\pm \sqrt{1-x^2}$

Since we are left with two answers for what $\cos (y)$ is equal to, we need to choose which one to use based on the domain of arcsin. You may remember from trigonometry that the domain of $\arcsin (x)$ is $x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, and on that domain of $x$, cosine is either always positive or equal to zero. So we will use the positive answer for $\cos (y)=\sqrt{1-x^2}$ and just plug that into the equation we found earlier to get the derivative in terms of $x$:

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

The process for finding the derivative of the inverse cosine or arccosine of $x$ is very similar so I will quickly go over the steps for that. Again let's let ${\cos }^{-1}(x)=y$ and $\cos (y)=x$. Then we can differentiate $\cos (y)=x$ implicitly with respect to $x$ to find the derivative of ${\cos }^{-1}(x)$ or $\arccos (x)$:

$\cos (y)=x$

$\frac{dy}{dx}[\cos (y)]=\frac{dy}{dx}[x]$

$-\sin (y)\cdot \frac{dy}{dx}=1$

$\frac{dy}{dx}=-\frac{1}{\sin (y)}$

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

Now that we have found the derivative of $\arcsin (x)$ to be equal to $\frac{1}{\sqrt{1-x^2}}$ and $\arccos (x)$ to be equal to $-\frac{1}{\sqrt{1-x^2}}$ there is one more trig function to differentiate, that being ${\tan }^{-1}(x)$ or $\arctan (x)$. We will follow a similar process where we let $\arctan (x)=y$ and $\tan (y)=x$ and then differentiate it implicitly:

$\tan (y)=x$

$\frac{dy}{dx}[\tan (y)]=\frac{dy}{dx}[x]$

${\sec }^2(y)\cdot \frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{{\sec }^2(y)}$

To find this in terms of $x$ we again use a pythagorean trigonometric identity. Specifically ${\tan }^2(y)+1={\sec }^2(y)$ and substitute $\tan (y)=x$ to get $x^2+1={\sec }^2(y)$. Now we just plug that in to get:

$\frac{dy}{dx}=\frac{1}{1+x^2}$

More on the Derivatives of the Inverse Trig Functions

Now that we know the derivatives of $\arcsin (x)$, $\arccos (x)$, $\arctan (x)$, and $\text{arccot}(x)$ you may be wondering about the derivatives of the reciprocal inverse functions. For the purposes of Calculus AB these derivatives are not needed to be known, they are also much more complex to derive so I won’t be doing that here. The derivatives of three main inverse trigonometric functions we just went over are very important to know though:

$\frac{dy}{dx}(\arcsin (x))=\frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}(\arccos (x))=-\frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}(\arctan (x))=\frac{1}{1+x^2}$

$\frac{dy}{dx}(\text{arccot}(x))=-\frac{1}{1+x^2}$

Applying the Derivatives of the Inverse Trig Functions

I am going to go over an example of a function where you need to know these derivatives to find the derivative of. That function is:$f(x)=e^{\arctan (2x)}$

I’ll go over the steps of differentiating it to quickly give you a feel as to how these derivatives are used.

$f’(x)=e^{\arctan (2x)}\cdot \frac{1}{1+(2x{)}^2}\cdot 2$

$f’(x)=\frac{2e^{\arctan (2x)}}{1+(2x{)}^2}$

$f’(x)=\frac{2e^{\arctan (2x)}}{1+4x^2}$

That's a short example of how you would use the derivative of $\arctan (x)$. Before the practice problems where you will apply these derivatives, let's go over some final notes for this topic.

Notes on the Derivatives of the Inverse Trig Functions

Practice Questions