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What does the derivative measure, and how do we differentiate powers and the standard functions?

Interpret the derivative as a gradient and rate of change, and differentiate powers of x and the standard exponential, logarithmic and trigonometric functions

A focused answer to the O-Level A-Maths outcome on basic differentiation. The meaning of the derivative as a gradient, the power rule, and the derivatives of the standard exponential, logarithmic and trigonometric functions.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to understand the derivative as the gradient of a curve and the instantaneous rate of change of one quantity with respect to another, and to differentiate powers of xx and the standard exponential, logarithmic and trigonometric functions. This is the entry point to the whole calculus strand.

The answer

What the derivative means

The derivative dydx\dfrac{dy}{dx} measures how fast yy changes as xx changes. Graphically it is the gradient of the tangent to the curve y=f(x)y = f(x) at a point. As a rate, it is the change in yy per unit change in xx at that instant.

The power rule

For any constant power nn:

ddxxn=nxn1.\frac{d}{dx}x^n = nx^{n-1}.

Multiply by the power, then reduce the power by one. A constant multiple stays attached (ddx(kxn)=knxn1\frac{d}{dx}(kx^n) = knx^{n-1}), and the derivative of a constant is zero. Differentiate a sum term by term.

Rewriting before differentiating

Roots and reciprocals must be rewritten as powers first: x=x1/2\sqrt{x} = x^{1/2} and 1x=x1\dfrac{1}{x} = x^{-1}. Then the power rule applies directly, and you tidy negative or fractional indices back at the end.

The standard derivatives

Three results to memorise:

ddxex=ex,ddxlnx=1x,ddxsinx=cosx,ddxcosx=sinx.\frac{d}{dx}e^x = e^x, \qquad \frac{d}{dx}\ln x = \frac{1}{x}, \qquad \frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x.

Note the minus sign when differentiating cosine. The exponential exe^x is the one function equal to its own derivative, which is why it appears so often in growth and decay.

Notation and the second derivative

The derivative is written dydx\dfrac{dy}{dx}, f(x)f'(x) or yy', all meaning the same thing. Differentiating a second time gives the second derivative d2ydx2\dfrac{d^2y}{dx^2} or f(x)f''(x), which measures how the gradient itself is changing and is used later to classify turning points.

Differentiating sums and multiples

Differentiation is linear: the derivative of a sum is the sum of the derivatives, and a constant multiple simply carries through. So a polynomial is differentiated term by term, each term handled by the power rule, with no need to expand brackets unless that makes a term a recognisable power.

Examples in context

Example 1. Speed from a distance graph. If distance ss is a function of time tt, the derivative dsdt\dfrac{ds}{dt} is the speed, the gradient of the distance-time graph, which is exactly how kinematics uses differentiation.

Example 2. Slope of a hillside. A cross-section of terrain modelled by y=f(x)y = f(x) has gradient dydx\dfrac{dy}{dx} giving the steepness at each point, so engineers read off where a slope is steepest by differentiating.

Try this

Q1. Differentiate y=x5y = x^5. [1 mark]

  • Cue. 5x45x^4.

Q2. Differentiate y=4xy = 4\sqrt{x}. [2 marks]

  • Cue. 412x1/2=2x4 \cdot \tfrac{1}{2}x^{-1/2} = \dfrac{2}{\sqrt{x}}.

Q3. Find dydx\dfrac{dy}{dx} for y=ex3cosxy = e^x - 3\cos x. [2 marks]

  • Cue. ex+3sinxe^x + 3\sin x.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksDifferentiate y=3x42x+xy = 3x^4 - \dfrac{2}{x} + \sqrt{x} with respect to xx.
Show worked answer →

Rewrite using indices: y=3x42x1+x1/2y = 3x^4 - 2x^{-1} + x^{1/2}.

Apply the power rule ddxxn=nxn1\dfrac{d}{dx}x^n = nx^{n-1} term by term:

dydx=12x32(1)x2+12x1/2=12x3+2x2+12x\dfrac{dy}{dx} = 12x^3 - 2(-1)x^{-2} + \dfrac{1}{2}x^{-1/2} = 12x^3 + \dfrac{2}{x^2} + \dfrac{1}{2\sqrt{x}}.

Markers reward rewriting as powers, the power rule on each term, and tidy positive-index form.

Original3 marksFind dydx\dfrac{dy}{dx} given y=2ex+5lnxsinxy = 2e^{x} + 5\ln x - \sin x.
Show worked answer →

Differentiate each standard function: ddxex=ex\dfrac{d}{dx}e^x = e^x, ddxlnx=1x\dfrac{d}{dx}\ln x = \dfrac{1}{x}, ddxsinx=cosx\dfrac{d}{dx}\sin x = \cos x.

So dydx=2ex+5xcosx\dfrac{dy}{dx} = 2e^x + \dfrac{5}{x} - \cos x.

Markers reward the three standard derivatives applied with their constant multiples and correct signs.

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