Section 1: Chain Rule

Theorem 1.1: Chain Rule

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then:

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

In Leibniz notation: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ where $u = g(x)$ .

Proof of Chain Rule:

Let $y = f(g(x))$ and $u = g(x)$ .

By definition:

\frac{dy}{dx} = \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h}

Let $\Delta u = g(x+h) - g(x)$ . Then:

\frac{dy}{dx} = \lim_{h \to 0} \frac{f(u + \Delta u) - f(u)}{\Delta u} \cdot \frac{\Delta u}{h} = f'(u) \cdot g'(x)

Therefore $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$ . ∎

∎

Example 1.1: Chain Rule Application

Find: $(\sin(x^2))'$

Solution: Let $u = x^2$ , then $y = \sin u$ .

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos u \cdot 2x = 2x\cos(x^2)

Example 1.2: Multiple Chain Rule

Find: $(e^{\sin(x^2)})'$

Solution: Apply chain rule twice:

\frac{d}{dx}e^{\sin(x^2)} = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x = 2x\cos(x^2)e^{\sin(x^2)}

Section 2: Inverse Function Derivatives

Theorem 2.1: Inverse Function Derivative

If $f$ is differentiable and has an inverse $f^{-1}$ , then:

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

Provided $f'(f^{-1}(y)) \neq 0$ .

Example 2.1: Derivative of Arctan

Prove: $(\arctan x)' = \frac{1}{1+x^2}$

Proof: Let $y = \arctan x$ , so $x = \tan y$ .

Differentiating: $1 = \sec^2 y \cdot \frac{dy}{dx}$

\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}

Section 3: Implicit Differentiation

Definition 3.1: Implicit Differentiation

When $y$ is defined implicitly by an equation $F(x, y) = 0$ , we differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Example 3.1: Circle Equation

Find: $\frac{dy}{dx}$ for $x^2 + y^2 = 25$

Solution: Differentiate both sides:

2x + 2y \frac{dy}{dx} = 0

Solving for $\frac{dy}{dx}$ :

\frac{dy}{dx} = -\frac{x}{y}

Example 3.2: Implicit with Exponential

Find: $\frac{dy}{dx}$ for $e^y = xy$

Solution: Differentiate:

e^y \frac{dy}{dx} = y + x \frac{dy}{dx}

Solving:

\frac{dy}{dx} = \frac{y}{e^y - x}

Section 4: Parametric Derivatives

Theorem 4.1: Parametric Derivative

If $x = x(t)$ and $y = y(t)$ are differentiable, then:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Provided $\frac{dx}{dt} \neq 0$ .

Example 4.1: Parametric Curve

Find: $\frac{dy}{dx}$ for $x = t^2, y = t^3$

Solution:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}

Section 5: Higher-Order Derivatives

Definition 5.1: Higher-Order Derivatives

The $n$ -th derivative of $f$ is defined recursively:

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

Notations: $f'', f''', f^{(n)}$ or $\frac{d^2y}{dx^2}, \frac{d^3y}{dx^3}, \frac{d^ny}{dx^n}$

Example 5.1: Second Derivative

Find: $f''(x)$ for $f(x) = e^{2x}$

Solution:

$f'(x) = 2e^{2x}$

f''(x) = 2 \cdot 2e^{2x} = 4e^{2x}

Example 5.2: n-th Derivative

Find: $f^{(n)}(x)$ for $f(x) = e^{ax}$

Solution: Each differentiation brings down a factor of $a$ :

f^{(n)}(x) = a^n e^{ax}

Section 6: Logarithmic Differentiation

Example 6.1: Logarithmic Differentiation

Find: $y'$ for $y = x^x$

Solution: Take logarithms: $\ln y = x \ln x$

Differentiate: $\frac{y'}{y} = \ln x + 1$

y' = y(\ln x + 1) = x^x(\ln x + 1)

Section 7: Related Rates

Example 7.1: Related Rates Problem

A balloon is inflated so its volume increases at 100 cm³/s. Find the rate of change of radius when r = 5 cm.

$V = \frac{4}{3}\pi r^3$ , so $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

At r = 5: $100 = 4\pi (25) \frac{dr}{dt}$

\frac{dr}{dt} = \frac{1}{\pi} \text{ cm/s}

Practice Quiz: Chain Rule and Higher Derivatives

10

Questions

0

Correct

0%

Accuracy

1

If

y = \sin(x^2)

, what is

\frac{dy}{dx}

?

Easy

Not attempted

2

If

y = e^{\sin x}

, find

\frac{dy}{dx}

.

Easy

Not attempted

3

For the inverse function, if

f'(x) = 3x^2

and

f(2) = 8

, what is

(f^{-1})'(8)

?

Medium

Not attempted

4

Using implicit differentiation on

x^2 + y^2 = 25

, find

\frac{dy}{dx}

.

Easy

Not attempted

5

If

x = t^2

and

y = t^3

, what is

\frac{dy}{dx}

?

Medium

Not attempted

6

What is the second derivative of

f(x) = e^{2x}

?

Easy

Not attempted

7

Find

(\ln(x^2+1))'

.

Easy

Not attempted

8

If

y = \arctan x

, what is

y'

?

Easy

Not attempted

9

For

e^y = xy

, find

\frac{dy}{dx}

using implicit differentiation.

Medium

Not attempted

10

What is

(\sin^2 x)'

?

Easy

Not attempted

Frequently Asked Questions

Why is the chain rule important?

The chain rule allows us to differentiate composite functions, which appear everywhere in applications. Without it, we couldn't differentiate functions like sin(x²), e^(cos x), or ln(1+x²).

How do I know when to use implicit differentiation?

Use implicit differentiation when y is not explicitly given as a function of x, such as in equations like x² + y² = 1 or xy + sin(y) = x.

What's the connection between parametric derivatives and chain rule?

The formula dy/dx = (dy/dt)/(dx/dt) comes from the chain rule: dy/dx = (dy/dt)·(dt/dx) = (dy/dt)/(dx/dt).

Is d²y/dx² the same as (dy/dx)²?

No! d²y/dx² is the second derivative (derivative of the derivative), while (dy/dx)² is the square of the first derivative. These are different quantities.

How do I identify the 'inner' and 'outer' functions?

The outer function is applied last. In f(g(x)), f is outer, g is inner. Example: sin(x²) has outer=sin, inner=x². Think about order of operations.