Topic 3.4 Notes – Differentiating Inverse Trigonometric Functions

The Derivatives of Inverse Trigonometric Functions

These derivatives come from the general inverse rule:

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

For trig functions, that rule gets simplified using identities like $\sin^2\theta + \cos^2\theta = 1$. You are not expected to re-derive them on a quiz or exam. Memorize the final forms.

The Six Formulas You Must Know

arcsin and arccos

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

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

• Same denominator
• arccos has the negative

arctan and arccot

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

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

• Same denominator
• arccot has the negative

arcsec and arccsc

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

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

• Absolute value is required
• arccsc has the negative

The Pattern (So You Don’t Memorize 6 Random Things)

There are three denominator types:

• $\sqrt{1-x^2}$ → arcsin, arccos
• $1+x^2$ → arctan, arccot
• $|x|\sqrt{x^2-1}$ → arcsec, arccsc

And in each pair, the second one is negative:

• cos⁻¹
• cot⁻¹
• csc⁻¹

If you remember that structure, it’s much harder to mix them up under time pressure.

Using the Chain Rule with Inverse Trig

Almost every test question wraps something inside the inverse trig function.

If

$$y = \sin^{-1}(g(x))$$

then

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

The inside function replaces every $x$ in the formula, then you multiply by its derivative.

Example

Let

$$y = \tan^{-1}(2x^3 - 5)$$

Derivative of arctan is $\frac{1}{1+x^2}$.

Replace $x$ with $2x^3 - 5$:

$$\frac{1}{1+(2x^3-5)^2}$$

Now multiply by derivative of the inside:

$$g'(x) = 6x^2$$

Final answer:

$$\frac{6x^2}{1+(2x^3-5)^2}$$

On FRQs, they love seeing whether you forget that last multiplication. If something is inside, there must be a chain rule factor.

Domain and Radical Awareness

The square roots tell you something about domain.

arcsin and arccos

Derivative contains $\sqrt{1-x^2}$.
Original domain is $-1 \le x \le 1$.

arcsec and arccsc

Derivative contains $\sqrt{x^2-1}$.
Defined when $|x| \ge 1$.

Why the absolute value?

In $\frac{1}{|x|\sqrt{x^2-1}}$, the absolute value ensures the denominator stays positive on both sides of the domain. Dropping it is a very common point deduction.

You usually won’t be asked to restate domain, but algebra mistakes under radicals show up a lot in multiple choice.

Common Mistakes That Lose Easy Points

• Forgetting the chain rule
• Dropping the negative on cos⁻¹, cot⁻¹, csc⁻¹
• Forgetting the absolute value in sec⁻¹ or csc⁻¹
• Writing $1 - g(x)^2$ incorrectly as $1 - g(x^2)$
• Mixing up $1-x^2$ and $1+x^2$

Quick mental trigger:

• arcsin / arccos → think circle identity → $1 - x^2$
• arctan / arccot → think Pythagorean identity → $1 + x^2$
• arcsec / arccsc → think restricted outside interval → absolute value