How to Differentiate Inverse Trig functions

What You Need to Know

Inverse trig derivatives show up constantly in AP Calc BC because they’re the “anti-trig-substitution” derivatives: when you see patterns like $\sqrt{1-x^2}$ or $1+x^2$ or $\sqrt{x^2-1}$, inverse trig often appears in antiderivatives and in derivative problems via chain rule.

Core idea

You’ll differentiate inverse trig functions by:

• Using the standard derivative formula (memorize it), then
• Applying the chain rule for the inside function $u(x)$.

If $y = \arcsin(u)$, you should immediately think:

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

Same idea for all six inverse trig functions.

Why it matters

• BC free response and MC love “nested” inverse trig like $\arctan(\sqrt{1-x})$.
• The absolute value in $\operatorname{arcsec}$ and $\operatorname{arccsc}$ derivatives is a classic trap.
• Sometimes you’re forced to use implicit differentiation to derive or verify formulas.

Step-by-Step Breakdown

The universal process (works every time)

1. Identify the outer function (which inverse trig?).
   • Example: $\arccos(\,\cdot\,)$ vs $\arctan(\,\cdot\,)$.
2. Set $u$ equal to the inside expression.
   • Example: if $y = \arcsin(3x^2-1)$, let $u = 3x^2-1$.
3. Write the correct inverse trig derivative template in terms of $u$.
   • For $\arcsin(u)$: $\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}$.
4. Differentiate the inside to get $u'$.
5. Substitute back and simplify carefully.
6. Domain/absolute value check (especially for \arcsec and \arccsc).

Mini worked walkthrough

Differentiate $y = \arctan(5x^3)$.

1. Outer: $\arctan(\cdot)$
2. Inside: $u=5x^3$
3. Template: $\frac{dy}{dx} = \frac{u'}{1+u^2}$
4. $u' = 15x^2$
5. Substitute: $y' = \frac{15x^2}{1+(5x^3)^2} = \frac{15x^2}{1+25x^6}$

When you might use implicit differentiation

If you forget a formula (or need to justify it), you can derive it:

• Start with $y = \arcsin(x) \Rightarrow \sin(y)=x$
• Differentiate implicitly, then solve for $\frac{dy}{dx}$ using a triangle or identity.

Key Formulas, Rules & Facts

Derivative formulas (memorize)

Chain rule reminder (do not skip)

If $y = f(u(x))$ then
$\frac{dy}{dx} = f'(u)\cdot u'$
Every inverse trig derivative formula above already includes $u'$—that’s the chain rule baked in.

Quick identity facts you’ll use during implicit derivations

• $1-\sin^2(y)=\cos^2(y)$
• $1+\tan^2(y)=\sec^2(y)$
• If $\sin(y)=x$ and $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, then $\cos(y)=\sqrt{1-x^2}$ (nonnegative on that interval)
• If $\sec(y)=x$, then $\tan^2(y)=\sec^2(y)-1=x^2-1$ and $|\tan(y)|=\sqrt{x^2-1}$, which leads to the absolute value in the derivative.

Critical warning: The $|u|$ in \frac{d}{dx}[\arcsec(u)] and \frac{d}{dx}[\arccsc(u)] comes from taking a square root when solving for $\tan(y)$ (or $\cot(y)$). Dropping it can make your derivative wrong for negative $u$.

Examples & Applications

Example 1: Basic chain rule with arcsin

Differentiate $y = \arcsin(3x^2-1)$.

• Let $u=3x^2-1$, so $u'=6x$
• Use arcsin formula:
  $y' = \frac{6x}{\sqrt{1-(3x^2-1)^2}}$
  Key insight: Keep the inside squared as a group—don’t expand unless asked.

Example 2: arccos with a quotient inside

Differentiate $y = \arccos\left(\frac{1}{x}\right)$.

• $u=\frac{1}{x}=x^{-1}$ so $u'=-x^{-2}=-\frac{1}{x^2}$
• arccos derivative has a minus:
  $y' = -\frac{u'}{\sqrt{1-u^2}} = -\frac{-\frac{1}{x^2}}{\sqrt{1-\left(\frac{1}{x}\right)^2}}$
  So
  $y' = \frac{\frac{1}{x^2}}{\sqrt{1-\frac{1}{x^2}}}$
  Optional simplification:
  $\sqrt{1-\frac{1}{x^2}}=\sqrt{\frac{x^2-1}{x^2}}=\frac{\sqrt{x^2-1}}{|x|}$
  Then
  $y' = \frac{\frac{1}{x^2}}{\frac{\sqrt{x^2-1}}{|x|}} = \frac{|x|}{x^2\sqrt{x^2-1}}$
  Key insight: Simplifying introduces $|x|$ sometimes—be careful.

Example 3: arctan with a radical inside

Differentiate $y = \arctan\left(\sqrt{1-x}\right)$.

• $u=(1-x)^{1/2}$
• $u' = \frac{1}{2}(1-x)^{-1/2}\cdot(-1)= -\frac{1}{2\sqrt{1-x}}$
• arctan rule:
  $y' = \frac{u'}{1+u^2} = \frac{-\frac{1}{2\sqrt{1-x}}}{1+(\sqrt{1-x})^2}$
  Since $(\sqrt{1-x})^2=1-x$,
  $y' = \frac{-\frac{1}{2\sqrt{1-x}}}{1+(1-x)} = \frac{-\frac{1}{2\sqrt{1-x}}}{2-x} = -\frac{1}{2\sqrt{1-x}(2-x)}$
  Key insight: Always simplify $1+u^2$—radicals often collapse nicely.

Example 4: arcsec (absolute value!)

Differentiate y = \arcsec(2x-1).

• $u=2x-1$ so $u'=2$
• arcsec derivative:
  $y' = \frac{2}{|2x-1|\sqrt{(2x-1)^2-1}}$
  Key insight: Do not drop $|2x-1|$. It matters when $2x-1<0$.

Common Mistakes & Traps

1. Forgetting the chain rule factor $u'$

   • What happens: You write $\frac{d}{dx}[\arcsin(3x)] = \frac{1}{\sqrt{1-(3x)^2}}$.
   • Why wrong: You ignored $u'=3$.
   • Fix: Always do “inside derivative” as your last check.

2. Missing the negative sign for $\arccos$ (and \arccot, \arccsc)

   • What happens: You treat $\arccos(u)$ like $\arcsin(u)$.
   • Why wrong: $\arccos$ derivative is $-\frac{u'}{\sqrt{1-u^2}}$.
   • Fix: Memorize: “cos and cot and csc are the negative ones” (details in memory aids).

3. Dropping absolute value in \arcsec / \arccsc derivatives

   • What happens: You write $\frac{u'}{u\sqrt{u^2-1}}$.
   • Why wrong: The correct denominator is $|u|\sqrt{u^2-1}$.
   • Fix: Put $|u|$ in first; only simplify later if you have domain info that guarantees $u>0$.

4. Mixing up $\sqrt{1-u^2}$ vs $\sqrt{u^2-1}$

   • What happens: You put $\sqrt{1-u^2}$ under arcsec.
   • Why wrong: Arcsec/arccsc relate to $\sec^2(y)-1=\tan^2(y)$, giving $u^2-1$.
   • Fix: Remember: “sec/csc go with $u^2-1$.”

5. Incorrect parentheses: squaring only part of the inside

   • What happens: $\sqrt{1-(3x^2-1)^2}$ becomes $\sqrt{1-3x^2-1^2}$ (nonsense).
   • Why wrong: You must square the entire inside expression.
   • Fix: Use big grouping: $1-(\text{inside})^2$.

6. Over-simplifying and creating invalid cancellations

   • What happens: You simplify $\sqrt{x^2}$ to $x$.
   • Why wrong: $\sqrt{x^2}=|x|$.
   • Fix: Any time you simplify a square root of a square, consider absolute value.

7. Assuming inverse trig derivatives “cancel” with trig functions

   • What happens: You think $\frac{d}{dx}[\arcsin(\sin x)] = 1$.
   • Why wrong: $\arcsin(\sin x)$ is not equal to $x$ for all $x$ (range restrictions!).
   • Fix: Only treat inverse trig as true inverses on the correct principal interval.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can write all six derivative formulas from memory:
  • \arcsin(u), \arccos(u), \arctan(u), \arccot(u), \arcsec(u), \arccsc(u)
• You always multiply by $u'$ (chain rule).
• You never forget the negatives for $\arccos$, \arccot, \arccsc.
• You use $\sqrt{1-u^2}$ for arcsin/arccos; $1+u^2$ for arctan/arccot; $\sqrt{u^2-1}$ for arcsec/arccsc.
• You include $|u|$ in arcsec/arccsc derivatives unless domain guarantees $u>0$.
• Your final answer has correct parentheses: $1-(\text{inside})^2$, not a mangled expansion.
• You’re careful when simplifying $\sqrt{x^2}$ into $|x|$.

You’ve got this—memorize the templates, then let the chain rule do the work.