Topic 3.4 Notes – Differentiating Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions

These are formulas you should be able to write from memory.

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

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

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

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

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

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

Patterns That Help You Remember

Instead of memorizing six random formulas, notice the structure:

If you remember the pattern, you can rebuild the chart quickly.

Why They Look Like This

All of these come from the inverse function rule:

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

For example, let $y = \sin^{-1}(x)$.
Rewrite it as $x = \sin(y)$.

Differentiate implicitly:

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

So

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

Then use the identity $\sin^2 y + \cos^2 y = 1$.
Since $\sin y = x$, we get:

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

That’s where the square root comes from.

You are not expected to redo this on a quiz. But knowing this helps you remember:

• Why there’s a square root
• Why signs matter
• Why absolute value shows up for sec and csc

Using the Chain Rule

On tests, inverse trig functions almost always appear inside something else.

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

then

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

That pattern works for all of them:

1. Differentiate the inverse trig function.
2. Leave the inside exactly as written.
3. Multiply by the derivative of the inside.

Example

Find the derivative of $y = \tan^{-1}(4x^3)$.

• Outer derivative → $\frac{1}{1 + (4x^3)^2}$
• Inside derivative → $12x^2$

Final answer:

$$y' = \frac{12x^2}{1 + 16x^6}$$

Notice the entire inside gets squared.

Absolute Value and Domain Details

The ones students forget most often:

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

That absolute value is required. It comes from domain restrictions of inverse secant.

If you have $\sec^{-1}(g(x))$, the denominator becomes:

$$|g(x)|\sqrt{(g(x))^2 - 1}$$

Dropping the absolute value can cost a point on an FRQ.

Where This Shows Up

You’ll see these in:

• Straight derivative questions (no calculator section)
• Composite functions that require the chain rule
• Implicit differentiation problems
• Related rates when trig relationships are inverted
• Inside FTC Part 1 expressions like
  $\frac{d}{dx}\left[\int_0^x \tan^{-1}(t^2)\,dt\right]$

If you see $1 + x^2$, think arctan.
If you see $\sqrt{1 - x^2}$, think arcsin or arccos.

Common Mistakes

• Forgetting the negative sign on arccos or arccot.
• Writing $1 - g(x^2)$ instead of $1 - (g(x))^2$.
• Forgetting to multiply by $g'(x)$.
• Dropping the absolute value in arcsec/arccsc.
• Squaring only part of the inside.