Differentiation Methods

Even a function that looks tangled has just a few tools for differentiating it. When one function sits inside another, the chain rule differentiates the outer part and then multiplies by the derivative of the inner part; when y is knotted in and cannot be solved for, implicit differentiation differentiates both sides as they stand; and when a curve moves along a time parameter t, parametric differentiation takes over. They look different, but underneath they share the same idea of multiplying and dividing small derivatives. Move the slider for x to see why the tangent slope of sin(2x) doubles, and increase the parameter t to watch the cycloid trace itself out.

Chain Rule for Composite Functions

x position1.5

🔗 The Chain Analogy

①In sin(2x), outer = sin, inner = 2x

②Differentiate outer first (cos), then multiply by inner derivative (2)

③Result: 2cos(2x) — that 2 is why the amplitude appears doubled

Chain Rule

{f(g(x))}' = f'(g(x)) · g'(x)

(derivative of outer) × (derivative of inner)

Implicit Differentiation

x² + y² = r² → 2x + 2y·y' = 0 → y' = −xy

Treat y as a function of x and differentiate both sides

🔄 Key Idea

①Works even when you can't write y = f(x)

②Apply the chain rule whenever y appears (dy/dx = y')

③Differentiate both sides, then solve for y'

Parametric Differentiation

Parameter t4

Parametric

x = f(t), y = g(t) → dydx = dy/dtdx/dt = g'(t)f'(t)

Differentiate each w.r.t. t, then divide dy/dt by dx/dt

🎡 Cycloid

①Trace of a point on a rolling wheel

②x = t − sin t, y = 1 − cos t

③Slope of the red tangent = (sin t)/(1 − cos t)

Logarithmic Differentiation

Log Differentiation

y = x^x → ln y = x ln x → y'/y = ln x + 1 → y' = x^x(ln x + 1)

Take ln of both sides; differentiating becomes much cleaner

💡 When to Use

①When the variable appears in both base and exponent (xˣ)

②When products and quotients are tangled

③Taking ln turns × into +, ÷ into −

Worked Examples

Example 1

Differentiate y = (2x+1)⁵.

Treat the outer function as u⁵ and the inner as u = 2x+1, then apply the chain rule.

y = (2x+1)⁵, u = 2x+1

Multiply the outer derivative 5u⁴ by the inner derivative (2x+1)′ = 2.

y' = 5(2x+1)⁴ · 2 = 10(2x+1)⁴

▸ y' = 10(2x+1)⁴

Chain rule = outer derivative × inner derivative. Never forget to multiply by the inner derivative.

Example 2

Find the slope of the tangent to x² + y² = 25 at the point (3, 4).

Differentiate both sides with respect to x (implicit differentiation).

2x + 2y·y' = 0 ⇒ y' = -xy

Substitute the point (3, 4).

y'|_(3,4) = -34

▸ -34

In implicit differentiation, multiply by y′ each time you differentiate y, then solve for y′ at the end.

Wrap-up

Three Methods

Chain: f'·g' Implicit: differentiate, solve for y' Parametric: g'(t)f'(t)

Chain rule, implicit, parametric — three core tools

Mock-exam style

For the curve given parametrically by x = t² + 1, y = t³ − t, what is dydx at t = 2?

①94

②52

③114

④3

⑤134

▸ ③ 114

Differentiate x and y with respect to t, then use dy/dx = (dy/dt)/(dx/dt).

dydx = 3t² - 12t

Substitute t = 2.

3(2)² - 12(2) = 114

🎯 Exam Points

①Chain rule: outer × inner — most common

②Implicit: don't forget to multiply by y' whenever y appears

③Parametric: dy/dx = (dy/dt)/(dx/dt)

④Log diff: for xˣ form, take ln then differentiate

⑤Mixed problems: be careful when chain + implicit are combined

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Derivatives of Special Functions

Applications of Derivatives

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