Differentiation Practice Questions

Q: When do I substitute into the derivative and when into the original equation?

Substitute into dydx\frac{dy}{dx}dxdy whenever you need a gradient, a rate, or the location of a turning point. Substitute into the original yyy when you need the actual yyy-coordinate of a point.

Q: How does the second-derivative test tell maximum from minimum?

Find d2ydx2\frac{d^{2}y}{dx^{2}}dx2d2y and put in the turning-point xxx. If it is positive, the point is a minimum; if it is negative, it is a maximum. A helpful picture: a valley curves upward (positive), a hill curves downward (negative).

Q: How do I know which rule to use, power, chain or something else?

Use the plain power rule ddx(axn)=anxn−1\frac{d}{dx}(ax^{n})=anx^{n-1}dxd(axn)=anxn−1 term by term for a polynomial. Reach for the chain rule only when a whole function sits inside a bracket raised to a power, such as (3x−1)4(3x-1)^{4}(3x−1)4; then differentiate the outside and multiply by the derivative of the inside.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from the plain power rule to a rate of change that links two derivatives. Give yourself roughly four to seven minutes per question and work on paper first, writing every line the way you would in the real exam, name the rule you are using, differentiate, then substitute the value of $x$ and state the gradient, point or rate that was asked for.

Resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, a correct derivative $\frac{dy}{dx}$ still earns method marks even when the final substitution slips, so always write the differentiated expression clearly before you put numbers in.

Two habits carry the whole chapter. First, $\frac{dy}{dx}$ is the gradient function: it gives the slope of the curve at any $x$ , so a gradient, a rate or a turning point always sends you to the derivative.

Second, the form of the answer matters, a gradient is a number, the equation of a tangent is a line, and a turning point is a coordinate together with its nature. Decide which of these the question is asking for before you write anything down.

Six practice questions

Q1[2 marks]

Given $y=3x^{4}-5x^{2}+7$ , find $\dfrac{dy}{dx}$ .

Show worked solution

Differentiate term by term with the power rule $\dfrac{d}{dx}(ax^{n})=anx^{\,n-1}$ : each term drops its power by one and multiplies by the old power. The constant $7$ differentiates to $0$ :

\frac{dy}{dx}=3(4)x^{3}-5(2)x^{1}+0

\frac{dy}{dx}=12x^{3}-10x

Answer

$\dfrac{dy}{dx}=12x^{3}-10x$ . Each term drops its power by one and multiplies by the old power; the lone constant vanishes.

Q2[3 marks]

Given $y=2x^{3}-4x+1$ , find the gradient of the curve at the point where $x=2$ .

Show worked solution

The gradient of a curve at a point is the value of $\dfrac{dy}{dx}$ there, so first differentiate to get the gradient function:

\frac{dy}{dx}=6x^{2}-4

Now substitute $x=2$ into the derivative, not into the original curve, which would give a $y$ -value instead of a gradient:

\left.\frac{dy}{dx}\right|_{x=2}=6(2)^{2}-4=6(4)-4=20

Answer

The gradient at $x=2$ is $20$ . Remember to substitute into the derivative, not into the original $y$ .

Q3[4 marks]

Given $y=(3x-1)^{4}$ , find $\dfrac{dy}{dx}$ and hence the gradient of the curve at $x=1$ .

Show worked solution

Because a whole function sits inside a power, use the chain rule: differentiate the outer power, keeping the inside unchanged, then multiply by the derivative of the inside $(3x-1)$ , which is $3$ :

\frac{dy}{dx}=4(3x-1)^{3}\times3=12(3x-1)^{3}

Now substitute $x=1$ . Work out the inside first, $3(1)-1=2$ , then cube it before multiplying:

\left.\frac{dy}{dx}\right|_{x=1}=12(2)^{3}=12(8)=96

Answer

$\dfrac{dy}{dx}=12(3x-1)^{3}$ , and the gradient at $x=1$ is $96$ . The factor of $3$ from the inside is the step most often forgotten.

Q4[5 marks]

A curve has equation $y=x^{2}-3x+4$ . Find the equation of (a) the tangent and (b) the normal to the curve at the point where $x=3$ .

Show worked solution

First find the point on the curve. Substitute $x=3$ into the equation: $y=(3)^{2}-3(3)+4=9-9+4=4$ , so the point is $(3,4)$ .

Then differentiate to get the gradient function and its value here:

\frac{dy}{dx}=2x-3 \;\Rightarrow\; \left.\frac{dy}{dx}\right|_{x=3}=2(3)-3=3

(a) The tangent has gradient $3$ and passes through $(3,4)$ . Use the straight-line form $y-y_{1}=m(x-x_{1})$ :

y-4=3(x-3) \;\Rightarrow\; y=3x-5

(b) The normal is perpendicular to the tangent, so its gradient is the negative reciprocal, $-\dfrac{1}{3}$ . Use the same point $(3,4)$ :

y-4=-\tfrac{1}{3}(x-3) \;\Rightarrow\; y=-\tfrac{1}{3}x+5

Answer

(a) Tangent: $y=3x-5$ . (b) Normal: $y=-\tfrac{1}{3}x+5$ , or equivalently $x+3y=15$ .

Check the point on each: at $x=3$ , the tangent gives $4$ and the normal gives $4$ .

Q5[5 marks]

Find the coordinates of the turning points of the curve $y=x^{3}-3x^{2}+4$ , and determine the nature of each.

Show worked solution

At a turning point the gradient is zero, so differentiate and set $\dfrac{dy}{dx}=0$ . Factorise to read off the $x$ -values:

\frac{dy}{dx}=3x^{2}-6x=3x(x-2)=0 \;\Rightarrow\; x=0 \text{ or } x=2

Find each $y$ by substituting back into the original curve: at $x=0$ , $y=4$ ; at $x=2$ , $y=8-12+4=0$ . Now differentiate again to test the nature with the second derivative:

\frac{d^{2}y}{dx^{2}}=6x-6

At $x=0$ : $6(0)-6=-6<0$ , so the curve is at a maximum. At $x=2$ : $6(2)-6=6>0$ , so the curve is at a minimum.

Answer

$(0,4)$ is a maximum point and $(2,0)$ is a minimum point. A positive second derivative means minimum; a negative one means maximum.

Q6[6 marks]

The radius $r\ \text{cm}$ of a circle increases at a constant rate of $0.3\ \text{cm s}^{-1}$ . Find (a) the rate of increase of the area when $r=4\ \text{cm}$ , (b) the rate of increase of the circumference.

[Area $A=\pi r^{2}$ , circumference $C=2\pi r$ .]

Show worked solution

(a) You are given a rate with respect to time and asked for another, so link them with the chain rule $\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}$ . Differentiate the area to get $\dfrac{dA}{dr}=2\pi r$ , and use the given $\dfrac{dr}{dt}=0.3$ :

\frac{dA}{dt}=2\pi r\times\frac{dr}{dt}=2\pi(4)(0.3)=2.4\pi

(b) Do the same for the circumference. Here $\dfrac{dC}{dr}=2\pi$ , a constant that does not depend on $r$ :

\frac{dC}{dt}=2\pi\times\frac{dr}{dt}=2\pi(0.3)=0.6\pi

Answer

(a) The area increases at $2.4\pi\approx7.54\ \text{cm}^{2}\,\text{s}^{-1}$ . (b) The circumference increases at $0.6\pi\approx1.88\ \text{cm s}^{-1}$ .

The circumference rate is constant because $\frac{dC}{dr}$ has no $r$ in it.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for a correct derivative, the right rule named, a clean substitution, and a final answer in the form the question asked for.

Go through the checklist below on every question, and be honest about the exact line where a mark was earned or lost.

Derivative mark: is $\frac{dy}{dx}$ correct, with each power dropped by one and multiplied by the old power?
Chain-rule mark: when a function sits inside a power, did you multiply by the derivative of the inside?
Substitution mark: did you put the $x$ -value into the derivative for a gradient, and into the original curve for the $y$ -coordinate?
Nature mark: for a turning point, did you test with the second derivative, positive means minimum, negative means maximum?
If your final number is wrong but the derivative and method are right, give yourself the method marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, forgetting the inside factor in the chain rule, substituting into $y$ instead of $\frac{dy}{dx}$ for a gradient, or misreading the sign of the second derivative, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

When do I substitute into the derivative and when into the original equation?

Substitute into $\frac{dy}{dx}$ whenever you need a gradient, a rate, or the location of a turning point. Substitute into the original $y$ when you need the actual $y$ -coordinate of a point.

How does the second-derivative test tell maximum from minimum?

Find $\frac{d^{2}y}{dx^{2}}$ and put in the turning-point $x$ . If it is positive, the point is a minimum; if it is negative, it is a maximum.

A helpful picture: a valley curves upward (positive), a hill curves downward (negative).

What is the gradient of a normal?

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient. If the tangent gradient is $3$ , the normal gradient is $-\frac{1}{3}$ .

How do I know which rule to use, power, chain or something else?

Use the plain power rule $\frac{d}{dx}(ax^{n})=anx^{n-1}$ term by term for a polynomial. Reach for the chain rule only when a whole function sits inside a bracket raised to a power, such as $(3x-1)^{4}$ ; then differentiate the outside and multiply by the derivative of the inside.

Does a wrong final answer cost me every mark?

No. Because marking is analytic, a correct derivative and a correct method still earn marks even if the arithmetic slips at the end.

Always write $\frac{dy}{dx}$ clearly before you substitute.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Differentiation, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

Ready to get started?