Differentiation and Applications
Differentiation and Applications
Differentiation questions cover the standard derivatives, the chain, product and quotient rules, implicit and parametric differentiation, tangents and normals, stationary points and their classification, rates of change and small increments. The applied parts of the paper also lean on this section for kinematics.
Rules to have automatic
- d/dx(xⁿ) = nx^(n−1); d/dx(e^(kx)) = ke^(kx); d/dx(ln x) = 1/x; d/dx(sin kx) = k cos kx; d/dx(cos kx) = −k sin kx; d/dx(tan x) = sec²x.
- Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v².
- Chain: dy/dx = dy/du × du/dx, used for nested functions and for connected rates of change.
- Parametric: dy/dx = (dy/dt)/(dx/dt). Implicit: differentiate every term, remembering that d/dx(y²) = 2y dy/dx.
Stationary points
Set dy/dx = 0. Classify with the second derivative: d²y/dx² < 0 maximum, > 0 minimum. If it is zero, use a sign table of dy/dx either side instead.
Worked example
y = x²e^(−x). Find the stationary points and classify them. Product rule with u = x², v = e^(−x): dy/dx = 2xe^(−x) − x²e^(−x) = x(2 − x)e^(−x). Since e^(−x) > 0, dy/dx = 0 at x = 0 and x = 2. Differentiating again: d²y/dx² = e^(−x)(x² − 4x + 2). At x = 0 this is 2 > 0, a minimum, with y = 0. At x = 2 it is e^(−2)(4 − 8 + 2) = −2e^(−2) < 0, a maximum, with y = 4e^(−2) ≈ 0.54.
Traps that cost marks
- Quotient rule with the numerator terms in the wrong order, flipping the sign.
- Forgetting the chain factor k in sin kx, cos kx and e^(kx).
- Losing the product term when differentiating implicitly y² or xy.
- Claiming a maximum or minimum when the second derivative is zero without checking signs.
- Using the gradient of the tangent where the normal was asked for.