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.

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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.

💡 It is best to memorize the following so you do not need to spend time re-deriving them during the test.

$f(x)$	$f'(x)$
$\frac{d}{dx}[\sin^{-1}(x)]$	$\frac{1}{\sqrt{1-x^{2}}}$
$\frac{d}{dx}[\cos^{-1}(x)]$	$-\frac{1}{\sqrt{1-x^{2}}}$
$\frac{d}{dx}[\tan^{-1}(x)]$	$\frac{1}{1+x^{2}}$
$\frac{d}{dx}[\csc^{-1}(x)]$	$-\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$
$\frac{d}{dx}[\sec^{-1}(x)]$	$\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$
$\frac{d}{dx}[\cot^{-1}(x)]$	$-\frac{1}{1+x^{2}}$

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

Forgetting the chain rule. 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 $\frac{3}{\sqrt{1-9x^2}}$ .
Mixing up the sign. The "co" functions ( $\cos^{-1}$ , $\cot^{-1}$ , $\csc^{-1}$ ) have a negative derivative. Dropping the negative is a common error.
Confusing the denominators. $\sin^{-1}$ and $\cos^{-1}$ use $\sqrt{1-x^2}$ , while $\tan^{-1}$ and $\cot^{-1}$ use $1+x^2$ . They are not interchangeable.
Dropping the absolute value. The derivatives of $\sec^{-1}(x)$ and $\csc^{-1}(x)$ include $|x|$ in the denominator. Leaving it out changes the answer.
Thinking inverse trig means reciprocal. $\sin^{-1}(x)$ means arcsine, not $\frac{1}{\sin(x)}$ . These are completely different functions with different derivatives.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
chain rule	A differentiation rule that provides a method for finding the derivative of a composite function by multiplying the derivative of the outer function by the derivative of the inner function.
derivative	The instantaneous rate of change of a function at a specific point, representing the slope of the tangent line to the function at that point.
inverse function	A function that reverses the effect of another function, such that if f(a) = b, then the inverse function f⁻¹(b) = a.
inverse trigonometric functions	Functions that reverse the action of trigonometric functions, such as arcsine, arccosine, and arctangent, which return an angle given a trigonometric ratio.

Frequently Asked Questions

What are inverse trig derivatives in AP Calculus?

Inverse trig derivatives are formulas for differentiating functions such as arcsin x, arccos x, and arctan x. In AP Calculus 3.4, you use these formulas with the chain rule when the inverse trig function has an inside function.

What is the derivative of arcsin x?

The derivative of arcsin x, also written sin inverse x, is 1 divided by the square root of 1 minus x squared. If the input is a function u, multiply by u prime using the chain rule.

What is the derivative of arctan x?

The derivative of arctan x, also written tan inverse x, is 1 divided by 1 plus x squared. For arctan of u, multiply that result by u prime.

Which inverse trig derivatives have negative signs?

The co-functions have negative signs: arccos x, arccot x, and arccsc x. Arcsin x, arctan x, and arcsec x use the corresponding positive forms.

How does the chain rule apply to inverse trig derivatives?

First identify the inverse trig formula for the outside function, then replace x with the inside expression and multiply by the derivative of that inside expression. This is the most common AP exam step students miss.

What is a common AP Calculus mistake with inverse trig derivatives?

A common mistake is confusing sin inverse notation with reciprocal sine. In AP Calculus, sin inverse x means arcsin x, not 1 over sin x, so use the inverse trig derivative formula.