← Differentiation methods

3 The rules & standard derivatives

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Differentiation rules

Power rule \frac{d}{dx}\left(x^n\right)=nx^{n-1} Chain rule \frac{d}{dx}f\bigl(g(x)\bigr)=f'\bigl(g(x)\bigr)\cdot g'(x) Differentiate the outside, keep the inside, times the derivative of the inside. Product rule (uv)′ = u′v + uv′ Quotient rule \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2} “Low d-high minus high d-low, over low squared.” ✕

Product rule vs chain rule — read the structure first

Product rule when two functions are MULTIPLIED: x²sin x. Chain rule when one function sits INSIDE another: sin(x²). Ask "is this A times B, or is this A of B?" before differentiating — many questions need both: x²sin(3x) is a product at the top level, with the chain rule applied to sin(3x) inside it.

Standard derivatives

Trig \frac{d}{dx}\sin x=\cos x\quad\frac{d}{dx}\cos x=-\sin x\quad\frac{d}{dx}\tan x=\sec^2 x Exponential \frac{d}{dx}e^{x}=e^{x}\qquad\frac{d}{dx}e^{kx}=ke^{kx} Logarithm \frac{d}{dx}\ln x=\frac{1}{x}\qquad\frac{d}{dx}\ln f(x)=\frac{f'(x)}{f(x)}

sec, cosec and cot

Reciprocal trig \frac{d}{dx}\sec x=\sec x\tan x\qquad\frac{d}{dx}\operatorname{cosec}x=-\operatorname{cosec}x\cot x\qquad\frac{d}{dx}\cot x=-\operatorname{cosec}^2x Each one is the quotient rule on 1/cos x, 1/sin x or cos x/sin x, so you can rebuild any of them in a line. The two that start with “co” carry a minus sign, as d/dx cos x does.

Showing that a function solves a differential equation

  • Differentiate y as many times as the equation needs: d²y/dx² means twice.
  • Substitute y and its derivatives into ONE side of the equation only, usually the left.
  • Simplify that side, line by line, until it is the other side exactly. Collect like terms and factorise to get there.
  • Finish with a sentence: “LHS = RHS, so y is a solution.” The marks are for the working being set out, not for the claim.
Worked example Chain + product together Differentiate y = x² e^(3x). Product rule with u = x², v = e^(3x). u′ = 2x. v′ = 3e^(3x) (chain rule on e^(3x)). y′ = u′v + uv′ = 2x·e^(3x) + x²·3e^(3x). = e^(3x)(2x + 3x²) = x e^(3x)(2 + 3x). ✓ dy/dx = x e^(3x)(3x + 2).

Practise just this