The rules, implicit differentiation and second derivatives · Differentiation

General Maths gave you the power, chain, product and quotient rules (see calculus). This lesson uses them on harder expressions, then on curves whose equation isn’t written as $y = \ldots$ at all.

The rules on harder expressions

Two habits make most questions short:

Simplify first. Write roots and fractions as powers of $x$ , or divide out, before differentiating: $\dfrac{3}{x^2} = 3x^{-2}$ .
Factorise the top after the quotient rule, so that a bracket cancels.

d/dx (u ⁄ v) = (v·u′ − u·v′) ⁄ v²

bottom × derivative of top, minus top × derivative of bottom, all over bottom squared

The quotient ruleBottom × top′ − top × bottom′, over bottom²

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q9 (a)

Differentiate $\dfrac{x^2 + 1}{(x + 1)^2}$ with respect to $x$ .

Name the parts
- ${u = x^2 + 1}$ , so ${u' = 2x}$ .
- ${v = (x + 1)^2}$ , so by the chain rule ${v' = 2(x + 1)}$ .
Think first. u is the top and v the bottom. What are u′ and v′?
Apply the rule
- The top is bottom × top′ minus top × bottom′: ${(x + 1)^2 \cdot 2x - (x^2 + 1) \cdot 2(x + 1)}$ .
- The bottom is the bottom squared: ${(x + 1)^4}$ .
Factorise the top
- Take out ${2(x + 1)}$ : the top is ${2(x + 1)[x(x + 1) - (x^2 + 1)]}$ .
- Simplify the bracket: ${x^2 + x - x^2 - 1 = x - 1}$ .
- Cancel one $(x + 1)$ : ${\frac{dy}{dx} = \frac{2(x - 1)}{(x + 1)^3}}$ .
Think first. Both terms on top share a factor. Which?

More: simplify first, then differentiate

Implicit differentiation

Some curves are given as an equation in $x$ and $y$ , such as $x^2 + y^2 = 25$ , and $y$ is hard or impossible to make the subject. Differentiate every term with respect to $x$ as it stands. A term in $y$ is differentiated as usual and then multiplied by $\frac{dy}{dx}$ , by the chain rule:

d/dx (y²) = 2y · dy/dx

d/dx (xy) = y + x · dy/dx

differentiate a y term as usual, then multiply by dy/dx; xy needs the product rule

Terms in yy² gives 2y dy/dx; xy needs the product rule

Then collect the $\frac{dy}{dx}$ terms on one side and factorise. The answer usually has both $x$ and $y$ in it, so you need both coordinates of the point.

Tangents to a curve in x and yPick a curve, slide the point

(2.7, 4.21)the point−0.642dy/dx = −x ÷ y

point = (2.7, 4.21)

Differentiate each term with respect to x: 2x + 2y·dy/dx = 0, so dy/dx = −x ÷ y. Put in the point's coordinates to get the gradient there: −0.642.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q1

If $x^2 - xy - y^2 = -31$ , find $\dfrac{dy}{dx}$ at $(2, 5)$ .

Differentiate term by term
- ${x^2}$ gives ${2x}$ .
- ${-xy}$ needs the product rule: ${-\left(y + x\frac{dy}{dx}\right)}$ .
- ${-y^2}$ gives ${-2y\frac{dy}{dx}}$ .
- The constant $-31$ gives 0.
- So ${2x - y - x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0}$ .
Think first. What do x², −xy and −y² each give?
Collect dy/dx
- ${2x - y = x\frac{dy}{dx} + 2y\frac{dy}{dx}}$ .
- Factorise: ${2x - y = (x + 2y)\frac{dy}{dx}}$ .
- So ${\frac{dy}{dx} = \frac{2x - y}{x + 2y}}$ .
Think first. Move the terms without dy/dx to the other side.
At (2, 5)
- ${\frac{dy}{dx} = \frac{4 - 5}{2 + 10} = -\frac{1}{12}}$ .
Think first. Put x = 2 and y = 5.

More: implicit differentiation

Second derivatives and “show that”

Differentiate twice to get $\dfrac{d^2y}{dx^2}$ . A “show that” question gives you the answer: work out each part separately, substitute, and simplify until the required result appears. Write each power of $x$ as $x^n$ first.

Worked example · WAEC 2012

WAEC 2012 · Paper 2 · Q10 (b)

Given that $y = \dfrac{px^2 + q}{x^4}$ , where $p$ and $q$ are constants, show that $x^2\dfrac{d^2y}{dx^2} + 7x\dfrac{dy}{dx} + 8y = 0$ .

Powers of x
- ${y = px^{-2} + qx^{-4}}$ .
Think first. Divide each term of the top by x⁴.
Differentiate twice
- ${\frac{dy}{dx} = -2px^{-3} - 4qx^{-5}}$ .
- ${\frac{d^2y}{dx^2} = 6px^{-4} + 20qx^{-6}}$ .
Each part of the left side
- ${x^2\frac{d^2y}{dx^2} = 6px^{-2} + 20qx^{-4}}$ .
- ${7x\frac{dy}{dx} = -14px^{-2} - 28qx^{-4}}$ .
- ${8y = 8px^{-2} + 8qx^{-4}}$ .
Think first. Multiply each derivative by what the question says.
Add them
- The $px^{-2}$ terms: ${6 - 14 + 8 = 0}$ .
- The $qx^{-4}$ terms: ${20 - 28 + 8 = 0}$ .
- So the left side is 0, as required.

More: second derivatives

Tangents and normals

The derivative at a point is the gradient $m$ of the tangent there. The normal is the line at right angles to the tangent, so its gradient is $-\frac1m$ :

Tangent and normalGradients multiply to −1: m × (−1/m) = −1

For either line you need a point and a gradient. Then use $y - y_1 = m(x - x_1)$ .

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q4

Find the equation of the tangent to the curve $y = \dfrac{1}{x + 1}$ when $x = 1$ .

The point
- ${y = \frac{1}{1 + 1} = \frac12}$ . The point is ${\left(1, \frac12\right)}$ .
Think first. Put x = 1 into the curve.
The gradient
- ${y = (x + 1)^{-1}}$ , so ${\frac{dy}{dx} = -(x + 1)^{-2}}$ .
- At $x = 1$ : ${-\frac{1}{4}}$ .
Think first. Write y as a power and use the chain rule.
The tangent
- ${y - \frac12 = -\frac14(x - 1)}$ .
- Multiply by 4: ${4y - 2 = -x + 1}$ .
- Rearrange: ${x + 4y - 3 = 0}$ .

More: tangents and normals

Stationary points

Setting $\frac{dy}{dx} = 0$ finds the points where the tangent is horizontal, and the second derivative tells a maximum from a minimum. FM12 covers these in full, with rates of change.

More: stationary points

Your turn

WAEC 2022 · Paper 2 · Q10 (b)

(b)
If $y = \dfrac{\sin x}{1 + \cos x}$ , find $\dfrac{dy}{dx}$ .

Worked solution (try it first)

(b)

Use the quotient rule with $u = \sin x$ and $v = 1 + \cos x$ : $u' = \cos x$ and $v' = -\sin x$ .
$\dfrac{dy}{dx} = \dfrac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2}$
$= \dfrac{\cos x + \cos^2x + \sin^2x}{(1 + \cos x)^2}$ .
Use $\cos^2x + \sin^2x = 1$ : the top is $1 + \cos x$ .
Cancel one factor: $\dfrac{dy}{dx} = \dfrac{1}{1 + \cos x}$ .

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