MATSEC IM Pure Mathematics study materials
MATSEC IM Pure Mathematics study resources

MATSEC IM Pure Mathematics: Differentiation

MATSEC IM Pure Mathematics · Work through your exam, one topic at a time.

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Differentiation

Differentiation

Differentiation questions test technique and interpretation: the derivative as a gradient, the chain, product and quotient rules, derivatives of exponential, logarithmic and trigonometric functions, and applications to tangents, normals, stationary points and rates of change.

Core rules

  • d/dx(x^n) = nx^(n−1); d/dx(e^x) = e^x; d/dx(ln x) = 1/x; d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = sec²x.
  • Chain rule: dy/dx = dy/du × du/dx. Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v².
  • Implicit: differentiate each term with respect to x, remembering that y-terms pick up a factor dy/dx.
  • Parametric: dy/dx = (dy/dt)/(dx/dt).
  • Stationary points occur where dy/dx = 0; classify by the sign of d²y/dx² (positive minimum, negative maximum) or by a sign table.

Worked example

Differentiate y = x²e^(3x). This is a product with u = x² and v = e^(3x). Then u′ = 2x and, by the chain rule, v′ = 3e^(3x). So dy/dx = 2x e^(3x) + x²(3e^(3x)) = x e^(3x)(2 + 3x). Factorising is worth a mark and makes the stationary points x = 0 and x = −2/3 obvious.

Traps

  • Differentiating e^(3x) as e^(3x) rather than 3e^(3x).
  • Reversing the order in the quotient rule, which flips the sign.
  • Forgetting dy/dx when differentiating a y-term implicitly.
  • Applying the second derivative test where it equals zero and drawing a conclusion anyway.