Inverse Trig Derivatives Practice Problems with Answers

These inverse trig derivatives practice problems start with the basic formulas, then use the chain rule, the product rule and a log, and finish with two problems that simplify in a surprising way and a tangent line. Try each one first. The answers come right after the list, then the full solutions.

The formulas you need:

Each formula comes from implicit differentiation. See the derivative of arcsin x, the derivative of arctan x and the derivative of arcsec x for the full proofs.

Problems

Basic formulas

1. y = 3\arctan x - 2\arcsin x
2. y = \arctan(3x)
3. y = \arccos(2x)

Chain rule

4. y = \arcsin(x^2)
5. y = \arctan\!\left(\sqrt{x}\right)
6. y = \arctan(e^x)
7. y = \operatorname{arcsec}(3x) for x > \frac{1}{3}

With other rules

8. y = x\arctan x
9. y = \ln(\arctan x) for x > 0
10. y = \arcsin x + \sqrt{1 - x^2}

Simplify and apply

11. y = \arctan x + \arctan\!\left(\dfrac{1}{x}\right) for x > 0
12. Find the tangent line to y = \arctan x at x = 1.

Answers

Step-by-step solutions

1. Use the table for each term:

2. With u = 3x, u' = 3 and u^2 = 9x^2:

3. With u = 2x, u' = 2 and u^2 = 4x^2. Keep the minus sign from arccos:

4. With u = x^2, u' = 2x and u^2 = x^4:

5. With u = \sqrt{x}, u' = \frac{1}{2\sqrt{x}} and u^2 = x:

6. With u = e^x, u' = e^x and u^2 = e^{2x}:

7. With u = 3x > 1, |u| = 3x and u' = 3:

8. Use the product rule:

9. Use (\ln u)' = \frac{u'}{u} with u = \arctan x:

10. The derivative of \sqrt{1 - x^2} is \frac{-2x}{2\sqrt{1 - x^2}} = -\frac{x}{\sqrt{1 - x^2}}. Add the two parts:

Since 1 - x^2 = (1 - x)(1 + x), this simplifies to \sqrt{\frac{1 - x}{1 + x}} for -1 < x < 1.

11. For the second term, u = \frac{1}{x} and u' = -\frac{1}{x^2}:

So y' = \frac{1}{1 + x^2} - \frac{1}{1 + x^2} = 0. The function is constant for x > 0. At x = 1 it equals \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}, so \arctan x + \arctan\frac{1}{x} = \frac{\pi}{2} for every x > 0.

12. The point is \arctan 1 = \frac{\pi}{4}. The slope is \frac{1}{1 + 1^2} = \frac{1}{2}:

Common mistakes on these problems

• Squaring only x, not the whole inside. In problem 2 the bottom is 1 + (3x)^2 = 1 + 9x^2, not 1 + 3x^2.
• Forgetting u' on top. \frac{d}{dx}\arcsin(x^2) needs the 2x in the numerator.
• Mixing up \arcsin x and \frac{1}{\sin x}. \arcsin x is the inverse function, and \frac{1}{\sin x} is \csc x. Their derivatives are completely different.
• Dropping the minus on arccos. Problem 3 has a negative answer because (\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the inverse trig derivatives lesson and the inverse function derivatives lesson. To review the regular trig functions first, use the trig derivatives practice problems.