3.4 Differentiating Inverse Trigonometric Functions

Differentiating inverse trig functions means knowing six derivative formulas and combining them with the chain rule. The cleanest way to find them is to apply the inverse function derivative formula, then use a trig identity like $\sin^2\theta + \cos^2\theta = 1$ to rewrite the result in terms of $x$. For AP Calculus, watch the negative signs on the co-functions: arccos, arccot, and arccsc.

Why This Matters for the AP Calculus Exam

Inverse trig derivatives show up in multiple-choice and free-response work, often mixed in with the chain rule, product rule, or quotient rule. The skill being tested is recognizing the form of a function and choosing the right procedure. Once you can spot $\sin^{-1}$, $\tan^{-1}$, and the others, you apply the matching formula and differentiate the inside. This builds directly on differentiating inverse functions and the chain rule, both heavily assessed in AP Calculus.

Key Takeaways

• The derivative of an inverse function comes from $\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$.
• You can derive every inverse trig derivative using that formula plus the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$.
• Memorize the six standard formulas so you do not waste time re-deriving them during the exam.
• Always apply the chain rule: multiply by the derivative of whatever is inside the inverse trig function.
• $\sin^{-1}$ and $\cos^{-1}$ derivatives share $\sqrt{1-x^2}$; $\tan^{-1}$ and $\cot^{-1}$ share $1+x^2$; $\sec^{-1}$ and $\csc^{-1}$ share $|x|\sqrt{x^2-1}$.
• The "co" versions ($\cos^{-1}$, $\cot^{-1}$, $\csc^{-1}$) just add a negative sign.

How to Find Derivatives of Inverse Trig Functions

You find the derivative of an inverse function by applying the chain rule with the definition of an inverse function, or by using the formula for the derivative of an inverse function:

$\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$

Here is how to apply that formula to find the derivative of inverse sine (arcsine).

Finding the Derivative of Inverse Sine

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

Start by applying the formula for the derivative of an inverse function:

$\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$

Since the derivative of $\sin(x)$ is $\cos(x)$, you get:

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

Now rewrite $\cos(y)$ in terms of $x$. By the definition of an inverse function, $x = \sin(y)$. Using the trig identity $\sin^{2}(y)+\cos^{2}(y)=1$, you can see that $\cos^{2}(y)=1-\sin^{2}(y)$.

Now simplify:

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

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

Plugging in $\cos(y)=\sqrt{1-x^{2}}$ gives:

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

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

The Derivatives of Inverse Trig Functions

You can run similar proofs to find the derivatives for the inverses of the other trig functions. That gives you the formulas below.

Notice the pattern: each "co" function ($\cos^{-1}$, $\cot^{-1}$, $\csc^{-1}$) is just the negative of its partner. Learning three formulas plus the negative sign rule covers all six.

How to Use This on the AP Calculus Exam

Problem Solving

When you spot an inverse trig function, identify which formula matches, then apply the chain rule by multiplying by the derivative of the inside.

Free Response

Show each step clearly: write the matching formula, substitute the inside function, and multiply by its derivative. Clear notation makes your work easy to follow and is important for clean exam work.

Question 1

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

Try solving it before checking 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,

$\frac{dy}{dx} = \frac{d}{dx}[\sin^{-1}(3x)]$

$= \frac{1}{\sqrt{1-(3x)^{2}}} \cdot \frac{d}{dx}[3x]$

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

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,

$\frac{dy}{dx} = \frac{d}{dx}[\tan^{-1}(x+6)]$

$= \frac{1}{1+(x+6)^{2}} \cdot \frac{d}{dx}[x+6]$

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

Common Misconceptions

• Misconception: The derivative of $\sin^{-1}(3x)$ is just $\frac{1}{\sqrt{1-(3x)^2}}$.
  Actually: The derivative of $\sin^{-1}(3x)$ is not $\frac{1}{\sqrt{1-(3x)^2}}$ by itself. You must multiply by the derivative of the inside, giving $\frac{3}{\sqrt{1-9x^2}}$.
• Misconception: The "co" functions ($\cos^{-1}$, $\cot^{-1}$, $\csc^{-1}$) have a positive derivative.
  Actually: The "co" functions ($\cos^{-1}$, $\cot^{-1}$, $\csc^{-1}$) have a negative derivative. Dropping the negative is a common error.
• Misconception: The denominators for inverse trigonometric functions are interchangeable.
  Actually: $\sin^{-1}$ and $\cos^{-1}$ use $\sqrt{1-x^2}$, while $\tan^{-1}$ and $\cot^{-1}$ use $1+x^2$. They are not interchangeable.
• Misconception: The derivatives of $\sec^{-1}(x)$ and $\csc^{-1}(x)$ do not include $|x|$ in the denominator.
  Actually: The derivatives of $\sec^{-1}(x)$ and $\csc^{-1}(x)$ include $|x|$ in the denominator. Leaving it out changes the answer.
• Misconception: $\sin^{-1}(x)$ means $\frac{1}{\sin(x)}$.
  Actually: $\sin^{-1}(x)$ means arcsine, not $\frac{1}{\sin(x)}$. These are completely different functions with different derivatives.

Related AP Calculus Guides

• 3.2 Implicit Differentiation
• 3.1 The Chain Rule
• 3.3 Differentiating Inverse Functions
• 3.6 Calculating Higher-Order Derivatives
• 3.5 Selecting Procedures for Calculating Derivatives
• Unit 3 Overview: Differentiation: Composite, Implicit, and Inverse Functions