How to Differentiate Inverse Trig functions

What You Need to Know

Inverse trig derivatives show up constantly in AP Calc AB because they’re the “antidotes” to trig functions when you’re differentiating compositions like \arcsin(3x) or \arctan(x^2). The big idea: **memorize the six derivative patterns**, then **apply the Chain Rule** when the inside isn’t just x.

Core idea (what you’re really doing)

If y = \arcsin(x), that means \sin(y)=x with y restricted to the principal range of arcsin. Differentiate implicitly:

\sin(y)=x
\cos(y)\,y' = 1
y' = \frac{1}{\cos(y)}

Then rewrite \cos(y) in terms of x using a triangle/Pythagorean identity. Since \sin(y)=x,

\cos(y)=\sqrt{1-\sin^2(y)}=\sqrt{1-x^2}

So,

\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}

That same implicit-diff + triangle idea produces all the inverse trig derivatives.

When/why you use these rules

• Any time you see y = \arcsin(u(x)), \arctan(u(x)), etc.
• In optimization/related rates when inverse trig pops up from geometry.
• On FRQs, they often test whether you can combine the inverse trig derivative with the Chain Rule and simplify correctly.

Critical reminder: \arcsin(x) does **not** mean \frac{1}{\sin(x)}. “Arc” means inverse function, not reciprocal.

Step-by-Step Breakdown

The fast, reliable process

1. Identify the outer function: which inverse trig function is it (arcsin, arccos, arctan, arcsec, etc.)?
2. Match the correct template for \frac{d}{dx} \,\text{arctrig}(u).
3. Multiply by the derivative of the inside (Chain Rule): if the inside is u(x), multiply by u'(x).
4. Simplify carefully:
   • Watch for missing negatives (especially arccos).
   • Watch for absolute values in arcsec/arccsc.
   • Simplify radicals/algebra cleanly.
5. (Optional but smart): Check domain restrictions if asked (or if simplification could hide them).

Mini worked walkthrough 1 (arcsin)

Differentiate y = \arcsin(3x).

• Template: \frac{d}{dx}\arcsin(u)=\frac{u'}{\sqrt{1-u^2}}
• Here u=3x so u'=3.

y' = \frac{3}{\sqrt{1-(3x)^2}} = \frac{3}{\sqrt{1-9x^2}}

Mini worked walkthrough 2 (arctan)

Differentiate y = \arctan(x^2+1).

• Template: \frac{d}{dx}\arctan(u)=\frac{u'}{1+u^2}
• u=x^2+1, so u'=2x.

y' = \frac{2x}{1+(x^2+1)^2}

Decision point: do you need absolute value?

• If you’re using arcsec or arccsc, the derivative includes |u|.
• For arcsin/arccos/arctan, you typically don’t introduce absolute value in the standard AB formulas.

Don’t “simplify away” the absolute value in \frac{1}{|u|\sqrt{u^2-1}} unless you’re explicitly told u>0 (or you’ve justified sign from context).

Key Formulas, Rules & Facts

The derivative templates (memorize these)

Why those square roots appear (quick intuition)

• arcsin/arccos come from \sin^2(y)+\cos^2(y)=1 leading to \sqrt{1-u^2}.
• arctan/arccot come from 1+\tan^2(y)=\sec^2(y) leading to 1+u^2.
• arcsec/arccsc come from \sec^2(y)-\tan^2(y)=1 leading to \sqrt{u^2-1} and the |u| from square-root sign issues.

Principal ranges (why signs/absolute values matter)

Knowing ranges helps explain the derivative forms and the sign choices.

• \arcsin(x) \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] so \cos(y)\ge 0 there.
• \arccos(x) \in \left[0,\pi\right] so \sin(y)\ge 0 there.
• \arctan(x) \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right).
• \arcsec(x) \in \left[0,\pi\right],\ y\ne \frac{\pi}{2} (common convention), which forces the absolute value in the derivative.

Chain Rule form you should default to

If y=\text{arctrig}(u(x)) then

y' = \left(\frac{d}{du}\text{arctrig}(u)\right)\cdot u'(x)

Write the template first, then plug in u and u'.

Examples & Applications

Example 1: arccos with a messy inside

Differentiate f(x)=\arccos(1-2x^3).

• Use \frac{d}{dx}\arccos(u)=-\frac{u'}{\sqrt{1-u^2}}.
• u=1-2x^3 so u'=-6x^2.

f'(x) = -\frac{-6x^2}{\sqrt{1-(1-2x^3)^2}} = \frac{6x^2}{\sqrt{1-(1-2x^3)^2}}

Key insight: two negatives make a positive; don’t drop the arccos negative.

Example 2: arcsin with algebra simplification

Differentiate g(x)=\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right).
Let u=\frac{x}{\sqrt{1+x^2}}=x(1+x^2)^{-1/2}.

Compute u':

u'=(1)(1+x^2)^{-1/2}+x\left(-\frac{1}{2}\right)(1+x^2)^{-3/2}(2x)
u'=(1+x^2)^{-1/2}-x^2(1+x^2)^{-3/2}
u' = \frac{1+x^2-x^2}{(1+x^2)^{3/2}} = \frac{1}{(1+x^2)^{3/2}}

Now use arcsin template:

g'(x)=\frac{u'}{\sqrt{1-u^2}}
Compute 1-u^2:

u^2=\frac{x^2}{1+x^2} \Rightarrow 1-u^2 = 1-\frac{x^2}{1+x^2}=\frac{1}{1+x^2}
So

\sqrt{1-u^2}=\sqrt{\frac{1}{1+x^2}}=\frac{1}{\sqrt{1+x^2}}

Therefore,

g'(x)=\frac{\frac{1}{(1+x^2)^{3/2}}}{\frac{1}{\sqrt{1+x^2}}}=\frac{1}{1+x^2}

Exam vibe: this looks nasty but collapses nicely if you simplify systematically.

Example 3: arctan with a quotient

Differentiate h(x)=\arctan\left(\frac{2x}{1-x^2}\right).
Let u=\frac{2x}{1-x^2}. Then

h'(x)=\frac{u'}{1+u^2}

Compute u' (quotient rule):

u' = \frac{(2)(1-x^2)-2x(-2x)}{(1-x^2)^2} = \frac{2-2x^2+4x^2}{(1-x^2)^2} = \frac{2+2x^2}{(1-x^2)^2} = \frac{2(1+x^2)}{(1-x^2)^2}

Now simplify 1+u^2:

1+u^2 = 1+\left(\frac{2x}{1-x^2}\right)^2 = \frac{(1-x^2)^2+4x^2}{(1-x^2)^2} = \frac{1+2x^2+x^4}{(1-x^2)^2}=\frac{(1+x^2)^2}{(1-x^2)^2}

So

h'(x)=\frac{\frac{2(1+x^2)}{(1-x^2)^2}}{\frac{(1+x^2)^2}{(1-x^2)^2}}=\frac{2}{1+x^2}

Key insight: with arctan, algebra often cancels beautifully if you combine into one fraction.

Example 4: arcsec (the absolute value trap)

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

• Template: \frac{d}{dx}\arcsec(u)=\frac{u'}{|u|\sqrt{u^2-1}}
• u=2x-1 so u'=2.

p'(x)=\frac{2}{|2x-1|\sqrt{(2x-1)^2-1}}

Do not drop the absolute value unless you have info about the sign of 2x-1.

Common Mistakes & Traps

1. Confusing inverse with reciprocal

   • Wrong: treating \arcsin(x) like \csc(x).
   • Why wrong: “arc” means inverse function.
   • Fix: say out loud: “arcsin is the angle whose sine is…”

2. Forgetting the Chain Rule (missing u')

   • Wrong: \frac{d}{dx}\arctan(5x)=\frac{1}{1+(5x)^2}.
   • Correct: multiply by 5: \frac{5}{1+25x^2}.
   • Fix: underline the inside function u(x) and compute u' every time.

3. Dropping the negative sign on arccos (and arccsc)

   • Wrong: \frac{d}{dx}\arccos(u)=\frac{u'}{\sqrt{1-u^2}}.
   • Correct: -\frac{u'}{\sqrt{1-u^2}}.
   • Fix: memorize the “cos is negative” fact (see Memory Aids).

4. Messing up the square root expression

   • Common error: writing \sqrt{1-u} instead of \sqrt{1-u^2}.
   • Fix: the arcsin/arccos denominator is always \sqrt{1-(\text{inside})^2}.

5. Ignoring the absolute value in arcsec/arccsc

   • Wrong: \frac{d}{dx}\arcsec(u)=\frac{u'}{u\sqrt{u^2-1}}.
   • Correct: \frac{u'}{|u|\sqrt{u^2-1}}.
   • Why: square roots produce nonnegative outputs; the sign of u matters.

6. Over-simplifying radicals in a way that changes domain

   • Example pitfall: replacing \sqrt{(x-2)^2} with x-2 instead of |x-2|.
   • Fix: remember \sqrt{a^2}=|a| for real a.

7. Not checking domain when asked (or when it matters)

   • arcsin/arccos require |u|\le 1.
   • arcsec/arccsc require |u|\ge 1.
   • Fix: if a problem asks “where is this differentiable?” combine the inside’s domain with where the denominator is nonzero.

8. Assuming inverse trig derivatives behave like trig derivatives

   • Wrong pattern: thinking \frac{d}{dx}\arctan(x)=\sec^2(x).
   • Correct: \frac{1}{1+x^2}.
   • Fix: inverse trig has its own set of templates.

Memory Aids & Quick Tricks

Quick Review Checklist

• Know all six templates:
  • \arcsin(u), \arccos(u), \arctan(u), \arccot(u), \arcsec(u), \arccsc(u)
• Always do Chain Rule: compute u'.
• arccos and arccsc derivatives have a negative sign.
• arcsin/arccos denominators: \sqrt{1-u^2}.
• arctan/arccot denominators: 1+u^2.
• arcsec/arccsc denominators: |u|\sqrt{u^2-1} (don’t drop |u|).
• If asked about domain/differentiability: enforce |u|\le 1 (arcsin/arccos) or |u|\ge 1 (arcsec/arccsc) and avoid zero denominators.
• Sanity-check: inverse trig derivatives should be algebraic expressions (no trig functions left) after simplification.

You’ve got this—if you can match the template and chain it cleanly, you’ll pick up these points fast.