Differentiation for NDA

Differentiation is a skill chapter. There is very little to understand once the rules are in place, and a lot to practise until they are automatic. NDA questions test the rules directly (a product, a chain, a quotient) and then in disguise: a function defined implicitly or through a parameter, a power like xˣ, or an inverse-trig expression that collapses after a substitution. The syllabus also names the derivative of one function with respect to another and second-order derivatives, and both turn up.

Typical question types are:

• a derivative using the product, quotient or chain rule;
• dy/dx from an implicit equation, often evaluated at a point;
• dy/dx for a curve given parametrically;
• derivatives of xˣ and similar forms, by taking logarithms;
• inverse-trig expressions that simplify with x = tan θ;
• the derivative of one function with respect to another;
• second derivatives, or showing a function satisfies an equation like y″ + 4y = 0.

What the derivative is

The derivative of f at x is the limit f′(x) = lim [f(x + h) − f(x)]/h as h → 0. It measures the instantaneous rate of change of f, which is also the slope of the tangent to the graph.

From first principles, for f(x) = x²: [(x + h)² − x²]/h = (2xh + h²)/h = 2x + h, which tends to 2x. You will rarely need first principles in the exam, but it explains why a function with a sharp corner, like |x| at 0, has no derivative there: the left and right limits of this ratio differ.

Standard derivatives

A pattern that halves the memorising: every "co" function (cos, cot, cosec, cos⁻¹, cot⁻¹) has a minus sign in its derivative.

The rules

• Sum and constant multiple: (u + v)′ = u′ + v′ and (ku)′ = ku′.
• Product: (uv)′ = u′v + uv′.
• Quotient: (u/v)′ = (u′v − uv′)/v².
• Chain: d/dx f(g(x)) = f′(g(x)) × g′(x). Differentiate the outside, keep the inside, then multiply by the derivative of the inside.

So d/dx sin 3x = 3 cos 3x, d/dx e raised to x² is 2x times e raised to x², and d/dx ln(sin x) = cos x/sin x = cot x.

Special techniques

Implicit differentiation. When y is tied up with x, differentiate both sides with respect to x and treat y as a function of x, so y² gives 2y dy/dx. Then solve for dy/dx. From x² + y² = 25: 2x + 2y dy/dx = 0, so dy/dx = −x/y.

Parametric differentiation. If x = f(t) and y = g(t), then dy/dx = (dy/dt) ÷ (dx/dt). For x = at² and y = 2at: dy/dx = 2a/(2at) = 1/t.

Logarithmic differentiation. For a variable raised to a variable power, or a long product, take ln first. If y = xˣ, then ln y = x ln x, so (1/y) dy/dx = ln x + 1 and dy/dx = xˣ(1 + ln x).

Derivative of one function with respect to another. The derivative of u with respect to v is (du/dx) ÷ (dv/dx).

Inverse-trig substitutions. Put x = tan θ to simplify expressions such as tan⁻¹[2x/(1 − x²)], sin⁻¹[2x/(1 + x²)] or cos⁻¹[(1 − x²)/(1 + x²)]. Each equals 2 tan⁻¹ x on the right interval, so each has derivative 2/(1 + x²) there.

Second derivatives. Differentiate twice: d²y/dx², also written y″. For y = x³, y′ = 3x² and y″ = 6x. The nth derivative of e raised to kx is kⁿ times e raised to kx.

Worked NDA-style MCQs

Q1. If y = xˣ, then dy/dx at x = 1 is:
(a) 0 (b) 1 (c) e (d) 2

dy/dx = xˣ(1 + ln x). At x = 1 this is 1 × (1 + 0) = 1. Answer: (b).

Q2. If x = at² and y = 2at, then dy/dx equals:
(a) t (b) 1/t (c) 2t (d) −1/t

dy/dt = 2a and dx/dt = 2at, so dy/dx = 1/t. Answer: (b).

Q3. For −1 < x < 1, the derivative of sin⁻¹[2x/(1 + x²)] is:
(a) 1/(1 + x²) (b) 2/(1 + x²) (c) 2/√(1 − x²) (d) −2/(1 + x²)

Put x = tan θ. Then 2x/(1 + x²) = sin 2θ, and for θ in (−π/4, π/4) the expression is 2θ = 2 tan⁻¹ x. Its derivative is 2/(1 + x²). Answer: (b).

Q4. The derivative of sin²x with respect to cos²x is:
(a) 1 (b) −1 (c) tan²x (d) −tan²x

d(sin²x)/dx = 2 sin x cos x and d(cos²x)/dx = −2 sin x cos x. Their ratio is −1. (Or notice that sin²x = 1 − cos²x.) Answer: (b).

Q5. For the curve x² + xy + y² = 7, dy/dx at (1, 2) is:
(a) −4/5 (b) 4/5 (c) −5/4 (d) −1

Check the point: 1 + 2 + 4 = 7. Differentiate: 2x + y + x dy/dx + 2y dy/dx = 0. At (1, 2): 2 + 2 + dy/dx + 4 dy/dx = 0, so dy/dx = −4/5. Answer: (a).

Q6. If y = A cos 2x + B sin 2x, then:
(a) y″ = 4y (b) y″ = −4y (c) y″ = −2y (d) y″ = 2y′

y′ = −2A sin 2x + 2B cos 2x and y″ = −4A cos 2x − 4B sin 2x = −4y. Answer: (b).

Common mistakes

• Missing the chain factor, especially with sin kx, e raised to kx, and ln of a function.
• Reversing the quotient rule numerator. It is u′v − uv′, with the derivative of the top first.
• Differentiating xˣ as x × xˣ⁻¹ or as xˣ ln x. Neither is right; take logarithms.
• Dividing the wrong way in parametric form. dy/dx is dy/dt over dx/dt.
• Forgetting that y is a function of x in implicit questions, so y² becomes 2y dy/dx, not 2y.

Practice set

1. d/dx(x³ + 3x²) is: (a) 3x² + 6x (b) 3x² + 3x (c) x² + 6x (d) 3x + 6
2. d/dx(cos 2x) is: (a) −sin 2x (b) 2 sin 2x (c) −2 sin 2x (d) −2 cos 2x
3. d/dx(ln x²) for x > 0 is: (a) 1/x² (b) 2/x (c) 2x (d) 1/(2x)
4. d/dx(x eˣ) is: (a) eˣ (b) x eˣ (c) eˣ(1 + x) (d) eˣ(x − 1)
5. d/dx ln(sin x) is: (a) cos x (b) tan x (c) cot x (d) 1/sin x
6. d/dx tan⁻¹(x²) is: (a) 1/(1 + x⁴) (b) 2x/(1 + x²) (c) 2x/(1 + x⁴) (d) 2/(1 + x⁴)
7. If y = √x + 1/√x, then 2x dy/dx + y equals: (a) 2√x (b) 2/√x (c) 0 (d) √x
8. The nth derivative of e²ˣ is: (a) e²ˣ (b) 2e²ˣ (c) 2ⁿe²ˣ (d) n e²ˣ

Answers:

1. (a). Power rule term by term.
2. (c). Chain factor 2.
3. (b). ln x² = 2 ln x.
4. (c). Product rule: eˣ + x eˣ.
5. (c). cos x/sin x.
6. (c). 1/(1 + x⁴) times the inner derivative 2x.
7. (a). dy/dx = 1/(2√x) − 1/(2x√x), so 2x dy/dx = √x − 1/√x. Adding y gives 2√x.
8. (c). Each differentiation brings down a factor of 2.

What to do next

• Write the standard-derivatives table from memory twice this week, including the inverse-trig rows.
• Solve 30 old NDA differentiation questions, marking which technique each needed.
• Revisit the limit definition in limits and continuity, then use these rules in applications of derivatives and run them backwards in integration.