Day 7: Chain Rule, Trig & Inverse Derivatives

The Chain Rule

Whenever you differentiate a composite function — a function stuffed inside another function — you need the chain rule. It's arguably the single most-used rule on the entire AP exam, since almost every 'real' problem involves some kind of composition.

Core Theorem

If

f(u)

is differentiable at

u=g(x)

and

g(x)

is differentiable at

x

, then the composite

f \circ g

is differentiable at

x

, and

(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

, or equivalently

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

Watch the TikTok ExplanationChain Rule Made Simple→

Step-by-Step SOP

1
Identify Outside and Inside
Decide which operation is applied last (the outside) and what it's being applied to (the inside).
2
Differentiate the Outside, Keep the Inside Unchanged
Apply the appropriate rule to the outside function, substituting the inside expression back in without simplifying it yet.
3
Multiply by the Inner Derivative
Multiply your result by the derivative of the inside function. If the inside itself is a composition, repeat the whole process on it.

Practice Exercises

Example 01Easy

Differentiate

y = (x^3 - 1)^{100}

.

NEED A HINT?

The 'outside' function is (something)

^{100}

and the 'inside' function is

x^3-1

.

SHOW DETAILED EXPLANATION

Step 1: Identify Outside and Inside

Outside:

u^{100}

. Inside:

u = x^3-1

.

Step 2: Differentiate the Outside, Keep the Inside

100(x^3-1)^{99}

.

Step 3: Multiply by the Derivative of the Inside

y' = 100(x^3-1)^{99} \cdot 3x^2 = 300x^2(x^3-1)^{99}

.

Example 02Medium

Differentiate

y = (2x+1)^5(x^3-2x+1)^4

.

NEED A HINT?

This needs BOTH the product rule and the chain rule — differentiate each factor with the chain rule, then combine with the product rule.

SHOW DETAILED EXPLANATION

Step 1: Differentiate Each Factor with the Chain Rule

\frac{d}{dx}(2x+1)^5 = 5(2x+1)^4 \cdot 2 = 10(2x+1)^4

.

\frac{d}{dx}(x^3-2x+1)^4 = 4(x^3-2x+1)^3(3x^2-2)

.

Step 2: Apply the Product Rule

y' = 10(2x+1)^4(x^3-2x+1)^4 + (2x+1)^5 \cdot 4(x^3-2x+1)^3(3x^2-2)

.

Step 3: Factor (Optional)

Both terms share

(2x+1)^4(x^3-2x+1)^3

, so this could be factored out — but on the AP exam, leaving it unsimplified like above still earns full credit.

Example 03Hard

Differentiate

h(x) = \sqrt[3]{x^2 + \sqrt{x^3+1}}

.

NEED A HINT?

This is a chain rule inside a chain rule — peel one layer at a time, from the cube root inward to the square root.

SHOW DETAILED EXPLANATION

Step 1: Rewrite as a Power

h(x) = \left(x^2 + (x^3+1)^{1/2}\right)^{1/3}

.

Step 2: Differentiate the Outermost Layer

h'(x) = \frac{1}{3}\left(x^2+(x^3+1)^{1/2}\right)^{-2/3} \cdot \frac{d}{dx}\left[x^2+(x^3+1)^{1/2}\right]

.

Step 3: Differentiate the Inner Expression (Another Chain Rule)

\frac{d}{dx}\left[x^2+(x^3+1)^{1/2}\right] = 2x + \frac{1}{2}(x^3+1)^{-1/2}(3x^2)

.

Step 4: Combine

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

.

Example 04AB/BC Standard

Let

F(x) = f(xf(f(x)))

. Given

f(1)=2, f(2)=3, f(3)=6, f'(1)=4, f'(2)=5, f'(3)=6

, find

F'(1)

.

NEED A HINT?

Work from the outside in, and at every step, evaluate the 'inside' expressions first so you know exactly which f-values and f'-values to use.

SHOW DETAILED EXPLANATION

Step 1: Differentiate the Outer Layer

Let

u(x) = xf(f(x))

, so

F(x) = f(u(x))

and

F'(x) = f'(u(x)) \cdot u'(x)

.

Step 2: Differentiate u(x) with the Product Rule

u(x) = x \cdot f(f(x)) \implies u'(x) = f(f(x)) + x \cdot f'(f(x))f'(x)

.

Step 3: Evaluate All Pieces at x = 1

f(1)=2

, so

f(f(1)) = f(2) = 3

. Then

u(1) = 1 \cdot f(f(1)) = 1 \cdot 3 = 3

. Also

f'(f(1)) = f'(2) = 5

and

f'(1)=4

.

Step 4: Compute u'(1)

u'(1) = f(f(1)) + 1 \cdot f'(f(1))f'(1) = 3 + (5)(4) = 23

.

Step 5: Compute F'(1)

F'(1) = f'(u(1)) \cdot u'(1) = f'(3) \cdot 23 = 6 \cdot 23 = 138

.

Common Pitfalls

⚠
Forgetting the Inner DerivativeThe most common chain rule mistake is stopping after differentiating the outside function and forgetting to multiply by the derivative of the inside — e.g. writing $\frac{d}{dx}(x^3-1)^{100} = 100(x^3-1)^{99}$ without the trailing $\cdot 3x^2$ .
⚠
Nested Compositions Need Multiple PassesWhen a function is nested three or four layers deep, you must apply the chain rule once per layer — peel one layer, multiply by its inner derivative, then repeat on what's left inside.

Derivatives of the Six Trigonometric Functions

The six trig derivatives should be memorized cold — once you have them down, differentiating any trig expression is just a matter of combining them with the chain, product, and quotient rules you already know.

Core Theorem

\frac{d}{dx}\sin x = \cos x

\frac{d}{dx}\cos x = -\sin x

\frac{d}{dx}\tan x = \sec^2 x

\frac{d}{dx}\cot x = -\csc^2 x

\frac{d}{dx}\sec x = \sec x \tan x

\frac{d}{dx}\csc x = -\csc x \cot x

Watch the TikTok ExplanationDerivative of Trigonometric Functions→

Step-by-Step SOP

1
Recall the Base Derivative
Identify which of the six trig functions is on the outside and recall its exact derivative, including sign.
2
Chain Rule for the Argument
If the argument isn't simply $x$ , multiply by the derivative of that argument — repeating for each additional layer of nesting.
3
Watch for Disguised Powers
Notation like $\sin^2 x$ or $\sec^3 x$ means the trig function is raised to a power — treat it as an outer power function wrapped around the trig function.

Practice Exercises

Example 01Easy

Differentiate

y = \sin(x^2)

.

NEED A HINT?

The outside function is

\sin(\cdot)

and the inside is

x^2

— apply the chain rule.

SHOW DETAILED EXPLANATION

Step 1: Differentiate sin(u), Keep u

\cos(x^2)

.

Step 2: Multiply by the Inner Derivative

y' = \cos(x^2) \cdot 2x = 2x\cos(x^2)

.

Example 02Medium

Differentiate

y = \sin^2 x

.

NEED A HINT?

Read

\sin^2 x

as

(\sin x)^2

— the outside function is (something)

^2

, and the inside is

\sin x

.

SHOW DETAILED EXPLANATION

Step 1: Rewrite Clearly

y = (\sin x)^2

.

Step 2: Differentiate the Outside, Keep the Inside

2(\sin x)^1

.

Step 3: Multiply by the Inner Derivative

y' = 2\sin x \cdot \cos x

(which can also be written as

\sin(2x)

using the double-angle identity).

Example 03Hard

Differentiate

g(t) = \tan(5 - \sin 2t)

.

NEED A HINT?

Two layers of chain rule: first peel off

\tan(\cdot)

, then peel off

\sin(\cdot)

inside it.

SHOW DETAILED EXPLANATION

Step 1: Differentiate the Outer tan Layer

\sec^2(5-\sin 2t) \cdot \frac{d}{dt}(5-\sin 2t)

.

Step 2: Differentiate the Inner Expression

\frac{d}{dt}(5-\sin 2t) = -\cos(2t) \cdot 2 = -2\cos 2t

(chain rule again on

\sin 2t

).

Step 3: Combine

g'(t) = -2\cos(2t)\sec^2(5-\sin 2t)

.

Example 04AB/BC Standard

Differentiate

f(x) = \sin(\cos(\tan x))

.

NEED A HINT?

Three layers deep — peel sin, then cos, then tan, multiplying by each inner derivative in turn.

SHOW DETAILED EXPLANATION

Step 1: Peel the Outermost sin Layer

\cos(\cos(\tan x)) \cdot \frac{d}{dx}[\cos(\tan x)]

.

Step 2: Peel the cos Layer

\frac{d}{dx}[\cos(\tan x)] = -\sin(\tan x) \cdot \frac{d}{dx}[\tan x] = -\sin(\tan x)\sec^2 x

.

Step 3: Combine All Layers

f'(x) = \cos(\cos(\tan x)) \cdot \left(-\sin(\tan x)\sec^2 x\right) = -\sec^2 x \sin(\tan x)\cos(\cos(\tan x))

.

Common Pitfalls

⚠
Cofunctions Carry a Negative Sign $\cos$ , $\cot$ , and $\csc$ (the 'co-' functions) all have negative derivatives, while $\sin$ , $\tan$ , and $\sec$ do not. Mixing up which three get the minus sign is a very common slip.
⚠
sin²x Is Still a Composite FunctionStudents often forget the chain rule applies to $\sin^2 x$ since it 'looks like' a basic trig function — remember it's really $(\sin x)^2$ , requiring the extra factor of $\cos x$ .

Derivatives of Inverse Functions

Sometimes you need the derivative of an inverse function $f^{-1}$ without ever solving for its formula explicitly — the inverse function theorem lets you get it directly from $f$ itself, just by finding one matching input/output pair.

Core Theorem

Inverse Function Theorem: if

f'(x)

exists on an interval

I

and is never

0

, then

f^{-1}

is differentiable, and

(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}

.

Watch the TikTok ExplanationDerivative of Inverse Functions→

Step-by-Step SOP

1
Find the Matching Input First
Solve $f(a) = b$ to identify $a = f^{-1}(b)$ before applying the inverse function theorem.
2
Apply (f⁻¹)'(b) = 1/f'(a)
Compute $f'(a)$ using ordinary differentiation rules, then take its reciprocal.
3
For Second Derivatives, Differentiate g'(x) Again
Treat $g'(x) = [f'(g(x))]^{-1}$ as its own function and apply the chain rule to get $g''(x)$ .

Practice Exercises

Example 01Medium

If

f(x) = x^3 - 2

, find

\frac{d}{dx}f^{-1}(x)\Big|_{x=6}

.

NEED A HINT?

First find the input

a

such that

f(a)=6

— that's

f^{-1}(6)

— then apply the inverse function theorem.

SHOW DETAILED EXPLANATION

Step 1: Find f⁻¹(6) by Solving f(a) = 6

a^3 - 2 = 6 \implies a^3 = 8 \implies a = 2

, so

f^{-1}(6) = 2

.

Step 2: Find f'(x) and Evaluate at a = 2

f'(x) = 3x^2 \implies f'(2) = 12

.

Step 3: Apply the Inverse Function Theorem

(f^{-1})'(6) = \frac{1}{f'(f^{-1}(6))} = \frac{1}{f'(2)} = \frac{1}{12}

.

Example 02Hard

Let

g

be the inverse of

f(x) = x^3+3x+1

. Find

g'(5)

and

g''(5)

.

NEED A HINT?

First find

g(5)

by solving

f(a)=5

(try small integers). Then find

g'(5)

from the inverse function theorem, and differentiate that formula again for

g''(5)

.

SHOW DETAILED EXPLANATION

Step 1: Find g(5) by Solving f(a) = 5

a^3+3a+1=5 \implies a^3+3a-4=0

. Testing

a=1

:

1+3-4=0

✓, so

g(5)=1

.

Step 2: Find f'(x) and f''(x)

f'(x) = 3x^2+3

, so

f'(1) = 6

.

f''(x) = 6x

, so

f''(1) = 6

.

Step 3: Find g'(5)

g'(5) = \frac{1}{f'(g(5))} = \frac{1}{f'(1)} = \frac{1}{6}

.

Step 4: Differentiate g'(x) = [f'(g(x))]⁻¹ Again for g''(x)

g''(x) = -\frac{f''(g(x))\,g'(x)}{[f'(g(x))]^2}

, so

g''(5) = -\frac{f''(1)\cdot g'(5)}{[f'(1)]^2} = -\frac{6\cdot\frac{1}{6}}{36} = -\frac{1}{36}

.

Common Pitfalls

⚠
You Don't Need the Explicit Inverse FormulaThe whole point of the inverse function theorem is to avoid solving for $f^{-1}(x)$ symbolically — just find the specific input value $a=f^{-1}(b)$ numerically, then use $f'(a)$ .
⚠
The Second Derivative Isn't Just 1/f''(a) $g''(x)$ requires differentiating $g'(x) = [f'(g(x))]^{-1}$ with the chain and power rules — it is NOT simply $\frac{1}{f''(a)}$ .

Derivatives of Inverse Trig Functions

The six inverse trig derivatives are a separate set of memorized formulas, built once from implicit differentiation and then reused forever — they show up constantly wrapped inside chain rule and product rule problems.

Core Theorem

\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}}

Step-by-Step SOP

1
Recall the Base Formula
Identify which of the six inverse trig derivatives applies, including its sign.
2
Chain Rule the Argument
Multiply by the derivative of whatever is inside the inverse trig function.
3
Combine with Product/Quotient Rules as Needed
If the inverse trig function is multiplied, divided, or raised to a power alongside other terms, layer in the product, quotient, or power rule around it.

Practice Exercises

Example 01Medium

Differentiate

f(x) = x\arctan(\sqrt{x})

.

NEED A HINT?

This needs both the product rule (for

x

times

\arctan(\cdot)

) and the chain rule (for

\sqrt{x}

inside the arctan).

SHOW DETAILED EXPLANATION

Step 1: Apply the Product Rule

f'(x) = \arctan(\sqrt{x}) + x\cdot\frac{d}{dx}\left[\arctan(\sqrt{x})\right]

.

Step 2: Differentiate arctan(√x) with the Chain Rule

\frac{d}{dx}\arctan(\sqrt{x}) = \frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}}

.

Step 3: Combine and Simplify

f'(x) = \arctan(\sqrt{x}) + \frac{x}{2\sqrt{x}(1+x)} = \arctan(\sqrt{x}) + \frac{\sqrt{x}}{2(1+x)}

.

Example 02Hard

Differentiate

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

.

NEED A HINT?

Rewrite as

[\sin^{-1}(x^2)]^{-1}

and use the chain rule, remembering the inverse sine derivative also needs its own chain rule for the

x^2

inside.

SHOW DETAILED EXPLANATION

Step 1: Rewrite as a Power

y = \left[\sin^{-1}(x^2)\right]^{-1}

.

Step 2: Differentiate the Outer Power

y' = -\left[\sin^{-1}(x^2)\right]^{-2}\cdot\frac{d}{dx}\left[\sin^{-1}(x^2)\right]

.

Step 3: Differentiate sin⁻¹(x²) with the Chain Rule

\frac{d}{dx}\sin^{-1}(x^2) = \frac{2x}{\sqrt{1-x^4}}

.

Step 4: Combine

y' = -\frac{2x}{\sqrt{1-x^4}\left[\sin^{-1}(x^2)\right]^2}

.

Common Pitfalls

⚠
Cofunction Derivatives Are Just Negatives $\cos^{-1}x$ , $\cot^{-1}x$ , and $\csc^{-1}x$ have exactly the same derivative formulas as $\sin^{-1}x$ , $\tan^{-1}x$ , and $\sec^{-1}x$ respectively, just with a minus sign — don't re-derive them from scratch.
⚠
Inverse Trig Arguments Still Need the Chain Rule $\frac{d}{dx}\sin^{-1}(x^2) \neq \frac{1}{\sqrt{1-x^4}}$ alone — you must still multiply by the derivative of the inside, $2x$ .

Identify Outside and Inside

Differentiate the Outside, Keep the Inside Unchanged

Multiply by the Inner Derivative

Practice Exercises

Step 1: Identify Outside and Inside

Step 2: Differentiate the Outside, Keep the Inside

Step 3: Multiply by the Derivative of the Inside

Step 1: Differentiate Each Factor with the Chain Rule

Step 2: Apply the Product Rule

Step 3: Factor (Optional)

Step 1: Rewrite as a Power

Step 2: Differentiate the Outermost Layer

Step 3: Differentiate the Inner Expression (Another Chain Rule)

Step 4: Combine

Step 1: Differentiate the Outer Layer

Step 2: Differentiate u(x) with the Product Rule

Step 3: Evaluate All Pieces at x = 1

Step 4: Compute u'(1)

Step 5: Compute F'(1)

Recall the Base Derivative

Chain Rule for the Argument

Watch for Disguised Powers

Practice Exercises

Step 1: Differentiate sin(u), Keep u

Step 2: Multiply by the Inner Derivative

Step 1: Rewrite Clearly

Step 2: Differentiate the Outside, Keep the Inside

Step 3: Multiply by the Inner Derivative

Step 1: Differentiate the Outer tan Layer

Step 2: Differentiate the Inner Expression

Step 3: Combine

Step 1: Peel the Outermost sin Layer

Step 2: Peel the cos Layer

Step 3: Combine All Layers

Find the Matching Input First

Apply (f⁻¹)'(b) = 1/f'(a)

For Second Derivatives, Differentiate g'(x) Again

Practice Exercises

Step 1: Find f⁻¹(6) by Solving f(a) = 6

Step 2: Find f'(x) and Evaluate at a = 2

Step 3: Apply the Inverse Function Theorem

Step 1: Find g(5) by Solving f(a) = 5

Step 2: Find f'(x) and f''(x)

Step 3: Find g'(5)

Step 4: Differentiate g'(x) = [f'(g(x))]⁻¹ Again for g''(x)

Recall the Base Formula

Chain Rule the Argument

Combine with Product/Quotient Rules as Needed

Practice Exercises

Step 1: Apply the Product Rule

Step 2: Differentiate arctan(√x) with the Chain Rule

Step 3: Combine and Simplify

Step 1: Rewrite as a Power

Step 2: Differentiate the Outer Power

Step 3: Differentiate sin⁻¹(x²) with the Chain Rule

Step 4: Combine

Gary Chang