Differentiation Techniques Notes (JC H2) — Free PDF

Differentiation techniques are the standard results and rules used to find derivatives in JC H2 Mathematics. This formula sheet lists the derivatives of powers, trigonometric, inverse trigonometric, exponential and logarithmic functions of f(x), together with the product rule, the quotient rule and implicit differentiation.

It also shows how to prove specific results such as d/dx (tan⁻¹ x) = 1 ÷ (1 + x²) and d/dx (aˣ) = aˣ ln a. It then shows how to manipulate derivatives to reach a target differential equation, a kind of variation that can surface in examinations.

What these notes cover

•Derivatives of axⁿ and (ax + b)ⁿ
•Derivative of [f(x)]ⁿ
•Trigonometric functions of f(x)
•Inverse trigonometric functions of f(x)
•Exponential and logarithmic functions
•Product rule and quotient rule
•Implicit differentiation
•Proving d/dx (tan⁻¹ x) = 1 ÷ (1 + x²)
•Proving d/dx (aˣ) = aˣ ln a
•Manipulating derivatives to obtain differential equations

Key formulas

Derivative of [f(x)]ⁿ

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

f'(x) is the derivative of f(x). With f(x) = ax + b this gives an(ax + b)ⁿ⁻¹.

Derivative of sin⁻¹[f(x)]

$\frac{d}{dx}\sin^{-1}[f(x)] = \frac{f'(x)}{\sqrt{1-[f(x)]^2}}$

For cos⁻¹[f(x)] the result carries a minus sign: −f'(x) ÷ √(1 − [f(x)]²).

Derivative of aᶠ⁽ˣ⁾

$\frac{d}{dx}\left[a^{f(x)}\right] = a^{f(x)}(\ln a)\,f'(x)$

For base e, d/dx [eᶠ⁽ˣ⁾] = f'(x) eᶠ⁽ˣ⁾. With f(x) = x, d/dx (aˣ) = aˣ ln a.

Product rule

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

u and v are functions of x.

Quotient rule

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

u and v are functions of x. Note the minus sign in the numerator.

Implicit differentiation

$\frac{d}{dx}[f(y)] = f'(y)\,\frac{dy}{dx}$

Use this to differentiate a function of y with respect to x, e.g. d/dx (tan y) = sec²y × dy/dx.

Worked example

If y = eˣ ln x, (a) find dy/dx; (b) show that x(d²y/dx²) + dy/dx − (1 + x)y = 2eˣ.

1(a) Using the product rule, dy/dx = eˣ(1 ÷ x) + eˣ ln x = eˣ ÷ x + y.
2(b) Differentiating again, d²y/dx² = eˣ ÷ x − eˣ ÷ x² + dy/dx.
3Multiplying through by x gives x(d²y/dx²) = eˣ − eˣ ÷ x + x(dy/dx).
4From (a), x(dy/dx) = eˣ + xy, so x(d²y/dx²) = 2eˣ − eˣ ÷ x + xy.
5Rearranging, x(d²y/dx²) + eˣ ÷ x − xy = 2eˣ.
6From (a), eˣ ÷ x = dy/dx − y, so x(d²y/dx²) + dy/dx − y − xy = 2eˣ.
7∴ x(d²y/dx²) + dy/dx − (1 + x)y = 2eˣ (shown).

Questions students ask

What is the derivative of tan⁻¹ x and how do you prove it?

d/dx (tan⁻¹ x) = 1 ÷ (1 + x²). Let y = tan⁻¹ x, so tan y = x, and differentiate both sides with respect to x to get sec²y (dy/dx) = 1. Then dy/dx = 1 ÷ sec²y = 1 ÷ (1 + tan²y) = 1 ÷ (1 + x²).

How do you differentiate a to the power of x?

d/dx (aˣ) = aˣ ln a. Let y = aˣ, so ln y = x ln a, and differentiate both sides with respect to x to get (1 ÷ y)(dy/dx) = ln a. Hence dy/dx = y ln a = aˣ ln a.

When do you use implicit differentiation in H2 Maths?

Use it to differentiate a function of y with respect to x, using d/dx [f(y)] = f'(y) × dy/dx. It is the method for proving the derivatives of tan⁻¹ x, sin⁻¹ x, cos⁻¹ x and aˣ. Exam questions may combine it with the product rule or the quotient rule.

What is the derivative of sec x and cosec x?

d/dx sec[f(x)] = f'(x) sec[f(x)] tan[f(x)] and d/dx cosec[f(x)] = −f'(x) cosec[f(x)] cot[f(x)]. With f(x) = x, these give sec x tan x and −cosec x cot x.

In the full PDF

•The complete two-page table of standard derivatives
•Derivatives of sin, cos, tan, cosec, sec and cot of f(x) side by side
•The remaining inverse trigonometric, exponential and logarithmic results, including tan⁻¹[f(x)] and ln f(x)
•Both proofs laid out line by line, with the note on proving sin⁻¹ x and cos⁻¹ x the same way
•The full 3-page formula sheet to print and keep for revision