Gradients and derivatives · Calculus (JAMB bridge)

The gradient of a curve

A straight line has the same gradient everywhere. A curve doesn’t: it is steep in some places and flat in others. The gradient of a curve at a point is the gradient of the tangent there, the straight line that just touches the curve at that point.

Gradient of a curveThe gradient of the tangent at the point: rise ÷ run

Chord to tangentAs Q slides towards P (h → 0), the chord's gradient becomes the tangent's

WAEC asks you to find a gradient from your graph: draw the tangent carefully at the point, choose two points far apart on it, and work out rise ÷ run.

Try it

The gradient of a curveSlide P along the curve

(1, 1)P2gradient of tangent

x at P = 1

The derivative is dy/dx = 2x, so at x = 1 the tangent's gradient is 2: the curve is going up.

Slide P and watch the gradient change. Switch on “Chord” and make $h$ small: the chord’s gradient closes in on the tangent’s. Switch on “Gradient graph” to see the gradient at every point plotted as a new curve.

The derivative

The gradient at any point is given by the derivative, written $\dfrac{dy}{dx}$ . For powers of $x$ there is one rule: bring the power down in front, then take 1 off the power.

y = ax^n \quad\Rightarrow\quad \frac{dy}{dx} = nax^{n-1}

A number on its own (a constant) has derivative 0, because its graph is a flat line. Differentiate a sum one term at a time.

d/dx (a xⁿ) = na x^{n − 1}

4x³ → 3 × 4x² = 12x²

bring the power down, then take 1 off the power; a constant gives 0

The power ruleBring the power down, then reduce the power by 1

Negative and fractional powers

Rewrite fractions and roots as powers first: $\frac1x = x^{-1}$ , $\frac{3}{x^2} = 3x^{-2}$ and $\sqrt x = x^{\frac12}$ . Then the same rule works. For example $\frac{d}{dx}\left(x^{-1}\right) = -x^{-2} = -\frac{1}{x^2}$ .

Simplify first

There is no rule yet for a product or a quotient, so expand brackets and split fractions over a single term before you differentiate: $\frac{2x^3 + x^2}{x} = 2x^2 + x$ .

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q9 (a)

Differentiate $y = 3t^4 - 4t^3 + 2t^2 - \dfrac1t - 5$ .

Rewrite 1/t as a power
$y = 3t^4 - 4t^3 + 2t^2 - t^{-1} - 5$ .
Think first. 1/t = t to what power?
Differentiate term by term
$12t^3 - 12t^2 + 4t - (-1)t^{-2} - 0$ .
Think first. Bring each power down and take 1 off.
Tidy up
$\frac{dy}{dt} = 12t^3 - 12t^2 + 4t + \frac{1}{t^2}$ .
Think first. What does −(−1)t⁻² become?

Using the gradient

The derivative gives the gradient at any point. It also works backwards: if you know the gradient, set $\frac{dy}{dx}$ equal to it and solve for $x$ . A tangent parallel to the $x$ -axis has gradient 0.

Worked example · JAMB 1994

JAMB 1994 · UME · Q40

Find the point $(x, y)$ where the curve $y = 2x^2 - 2x + 3$ has gradient 2.

The gradient function
$\frac{dy}{dx} = 4x - 2$ .
Think first. Differentiate 2x² − 2x + 3.
Set it equal to 2
$4x = 4$ , so $x = 1$ .
Think first. Solve 4x − 2 = 2.
Find y
$y = 2 - 2 + 3 = 3$ . The point is $(1, 3)$ : option A.
Think first. Put x = 1 into the curve.

Your turn

JAMB 1997 · UME · Q40

Calculus (JAMB bridge)

Find the gradient of the curve $y = 2\sqrt x - \frac1x$ at the point $x = 1$ .

Worked solution (try it first)

Write the terms as powers: $y = 2x^{1/2} - x^{-1}$ .
Differentiate: $\frac{dy}{dx} = x^{-1/2} + x^{-2}$ .
At $x = 1$ : $1 + 1 = 2$ , option C.

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