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)

Identify the outer function (which inverse trig?).
- Example: \arccos(\,\cdot\,) vs \arctan(\,\cdot\,).
Set u equal to the inside expression.
- Example: if y = \arcsin(3x^2-1), let u = 3x^2-1.
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}}.
Differentiate the inside to get u'.
Substitute back and simplify carefully.
Domain/absolute value check (especially for \arcsec and \arccsc).

Mini worked walkthrough

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

Outer: \arctan(\cdot)
Inside: u=5x^3
Template: \frac{dy}{dx} = \frac{u'}{1+u^2}
u' = 15x^2
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)

Function y	Derivative \frac{dy}{dx}	When to use	Notes / traps
\arcsin(u)	\frac{u'}{\sqrt{1-u^2}}	Inside looks like \sqrt{1-u^2} patterns	Domain of arcsin: u\in[-1,1] (derivative blows up at \pm1)
\arccos(u)	-\frac{u'}{\sqrt{1-u^2}}	Similar to arcsin but negative	Easy sign trap: arccos has the minus
\arctan(u)	\frac{u'}{1+u^2}	Inside looks like 1+u^2	No square roots; usually “clean”
\operatorname{arccot}(u)	-\frac{u'}{1+u^2}	Less common; sometimes appears in old-style problems	Many courses treat arccot differently by convention; AP Calc standard uses the negative form
\operatorname{arcsec}(u)	\frac{u'}{\|u\|\sqrt{u^2-1}}	Inside looks like \sqrt{u^2-1}	Absolute value is required; domain: \|u\|\ge1
\operatorname{arccsc}(u)	-\frac{u'}{\|u\|\sqrt{u^2-1}}	Same pattern as arcsec but negative	Minus + absolute value

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

Common Mistakes & Traps

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.
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).
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.
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.”
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.
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.
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

Trick / mnemonic	What it helps you remember	When to use
“SCA” are the negatives: \arccos, \arccot, \arccsc	Which inverse trig derivatives start with a minus sign	Quick sign check before final answer
“sin/cos live in 1-u^2; tan lives in 1+u^2; sec/csc live in u^2-1”	Which expression goes under the radical / in denominator	Picking the right template fast
Arcsec/arccsc: “ABS + radical”	You must have \|u\|\sqrt{u^2-1} in the denominator	Anytime you see \arcsec or \arccsc
If inside is messy, set u= inside and compute u' on the side	Prevents missing chain rule and reduces algebra mistakes	Nested expressions like \arctan(\sqrt{1-x})
Don’t expand 1-u^2 unless forced	Avoids algebra errors and wasted time	Most multiple-choice and FRQ simplifications

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.