3.4 Differentiating Inverse Trigonometric Functions

Now that we have learned in the previous section 3.3 Differentiating Inverse Functions how to differentiate inverse functions, we can apply that knowledge to inverse trigonometric functions and find their derivatives, too!

🔍 How to Find Derivatives of Inverse Trig Functions

Recall that we find the derivative of an inverse function by applying the chain rule with the definition of an inverse function or the formula for the derivative of an inverse function:

Here is how we can apply the formula for the derivative of an inverse function to find the derivative of inverse sine or arcsine!

🤔 Finding the Derivative of Inverse Sine

If $y = \sin^{-1}(x)$, what is $\frac{dy}{dx}$?

We start by applying the formula for the derivative of an inverse function:

Since the derivative of $\sin(x)$ is $\cos(x)$, we can determine that…

Then, rewriting cos(y) in terms of x, we get $x = \sin(y)$, by the definition of an inverse function. And using the trig identity $\sin^{2}(y)+\cos^{2}(y)=1$, we can see that $\cos^{2}(y)=1-\sin^{2}(y)$.

Now, start simplifying!

We’re almost done! Therefore, by plugging in $\cos(y)=\sqrt{1-x^{2}}$, we know that…

Finally, the derivative of $\sin^{-1}(x)$ is $\frac{1}{\sqrt{1-x^{2}}}$.

📚 The Derivatives of Inverse Trig Functions

We can do similar proofs as the one above to find the derivatives for the inverses of the other trig functions. This will get us the following derivatives.

📝 Differentiating Inverse Trig Functions Practice

Now it’s time to practice what you’ve learned!

Question 1

If $y = \sin^{-1}(3x)$, what is $\frac{dy}{dx}$?

Try solving it before taking a look at the answer below!

Answer: $\frac{3}{\sqrt{1-9x^{2}}}$

Solution:

The formula for the derivative of $\sin^{-1}(x)$ is $\frac{d}{dx}[\sin^{-1}(x)] =$ $\frac{1}{\sqrt{1-x^{2}}}$. (see chart above)

Using the chain rule,

Question 2

If $y = \tan^{-1}(x+6)$, what is $\frac{dy}{dx}$?

Answer: $\frac{1}{x^{2}+12x+37}$

Solution:

The formula for the derivative of $\tan^{-1}(x)$ is $\frac{d}{dx}[\tan^{-1}(x)] =$ $\frac{1}{1+x^{2}}$. (see chart above)

Using the chain rule,

Great work! You did it!