Differentiation Worked Examples (KBAT)

Q: How do I know which chain to build for a rate of change?

Write the rate you want and the rate you are given, then connect them through a shared variable. For yyy and xxx with xxx changing in time, that link is dydt=dydx×dxdt\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}dtdy=dxdy×dtdx. Evaluate dydx\frac{dy}{dx}dxdy at the stated instant.

Q: Why does the small-change formula use the gradient at the starting point?

The approximation δy≈dydx δx\delta y\approx\frac{dy}{dx}\,\delta xδy≈dxdyδx treats the curve as a straight line over the tiny interval. That line has the gradient at where you start, so you evaluate dydx\frac{dy}{dx}dxdy at the initial xxx-value before multiplying by δx\delta xδx.

Q: In an optimisation problem, why must I get down to one variable?

You can only differentiate with respect to a single variable. Use the constraint, here the fixed 242424 m of fencing, to write one variable in terms of the other, so the quantity you are maximising becomes a function of just xxx.

Q: Do I always need the second derivative to confirm a maximum?

It is the cleanest confirmation and is expected for full marks. If d2Adx2<0\frac{d^{2}A}{dx^{2}}<0dx2d2A<0 the stationary point is a maximum; if it were positive, a minimum. A sign check on dAdx\frac{dA}{dx}dxdA either side also works.

What these examples cover

These hard Differentiation examples combine ideas the way the exam does. You will handle a rate of change by chaining derivatives, $\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}$ ; estimate a small change with $\delta y\approx\frac{dy}{dx}\,\delta x$ ; and solve an optimisation problem where the first job is to build a function of one variable, then differentiate to maximise it.

The numbers stay clean so the reasoning is never lost in arithmetic. Cover the solution, attempt the whole question on paper, then check line by line.

On these longer questions, the marks live in the setup, the relation you write down before you differentiate, so read each one twice before starting.

Worked examples

Work through all three. Each begins with a decision about what to differentiate and with respect to what, get that right and the calculus that follows is short.

Q1[5 marks]

Two variables are related by $y=\dfrac{4}{x}+3x$ . Given that $x$ increases at a constant rate of $2$ units per second, find the rate of change of $y$ at the instant when $x=2$ .

Show worked solution

A rate of change in time uses the chain rule $\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}$ . We are told $\frac{dx}{dt}=2$ ; we need $\frac{dy}{dx}$ at $x=2$ .

Rewrite the reciprocal as a power first:

y=4x^{-1}+3x \;\Rightarrow\; \frac{dy}{dx}=-4x^{-2}+3=-\frac{4}{x^{2}}+3

Evaluate the gradient at $x=2$ :

\left.\frac{dy}{dx}\right|_{x=2}=-\frac{4}{(2)^{2}}+3=-1+3=2

Now multiply by $\frac{dx}{dt}$ :

\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}=2\times 2=4

Answer

$y$ is increasing at $4$ units per second when $x=2$ . The whole method is: differentiate to get $\frac{dy}{dx}$ , evaluate it at the given instant, then multiply by the given $\frac{dx}{dt}$ .

Q2[4 marks]

Given $y=x^{3}-2x$ , use differentiation to find the approximate change in $y$ when $x$ increases from $3$ to $3.02$ . Hence estimate the value of $y$ at $x=3.02$ .

Show worked solution

A small change is estimated with $\delta y\approx\frac{dy}{dx}\,\delta x$ . Differentiate and evaluate the gradient at the starting value $x=3$ :

\frac{dy}{dx}=3x^{2}-2 \;\Rightarrow\; \left.\frac{dy}{dx}\right|_{x=3}=3(3)^{2}-2=27-2=25

The change in $x$ is $\delta x=3.02-3=0.02$ . Substitute both into the approximation:

\delta y\approx\frac{dy}{dx}\,\delta x=25\times 0.02=0.5

For the estimated value, add $\delta y$ to $y$ at $x=3$ , where $y=(3)^{3}-2(3)=27-6=21$ :

y(3.02)\approx 21+0.5=21.5

Answer

The approximate change is $\delta y\approx 0.5$ , so $y\approx 21.5$ at $x=3.02$ . The exact value is $3.02^{3}-2(3.02)=21.5036$ , so the estimate is close, as expected for a small $\delta x$ .

Q3[7 marks]

A rectangular plot is to be enclosed on three sides by fencing, with a long existing wall forming the fourth side. There are $24$ m of fencing available.

Find the dimensions that give the greatest possible area, and state that maximum area, confirming it is a maximum.

Show worked solution

Let the two sides perpendicular to the wall each be $x$ m, and the side parallel to the wall be $y$ m. Only three sides are fenced, so the fencing used is:

2x+y=24 \;\Rightarrow\; y=24-2x

Write the area as a function of $x$ alone by substituting for $y$ :

A=xy=x(24-2x)=24x-2x^{2}

For the greatest area, differentiate and set the derivative to zero:

\frac{dA}{dx}=24-4x=0 \;\Rightarrow\; x=6

Confirm this is a maximum with the second derivative, then find $y$ and the area:

\frac{d^{2}A}{dx^{2}}=-4<0 \;\;(\text{maximum})

y=24-2(6)=12, \qquad A=6\times 12=72

Answer

The plot should be $6$ m by $12$ m, giving a maximum area of $72\ \text{m}^{2}$ . The second derivative is $-4$ , negative, so the stationary value is indeed a maximum, not a minimum.

Q4[5 marks]

The curve $y = x^{2} - 3x - 4$ passes through the point $A(2,-6)$ . Find the equation of the normal to the curve at $A$ .

This normal cuts the $x$ -axis at $P$ and the $y$ -axis at $Q$ . Find the area of triangle $OPQ$ , where $O$ is the origin.

Show worked solution

Differentiate to find the gradient function, then evaluate it at $x=2$ to get the gradient of the tangent at $A$ .

\frac{dy}{dx}=2x-3 \;\Rightarrow\; \left.\frac{dy}{dx}\right|_{x=2}=2(2)-3=1

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal. Use this with point $A(2,-6)$ to write the equation of the normal:

m_{\text{normal}}=-\frac{1}{1}=-1 \;\Rightarrow\; y-(-6)=-1(x-2) \;\Rightarrow\; y=-x-4

Find where this line meets each axis: set $y=0$ for $P$ , and $x=0$ for $Q$ .

P: 0=-x-4 \Rightarrow x=-4,\;(-4,0) \qquad Q: x=0 \Rightarrow y=-4,\;(0,-4)

\text{Area}=\frac{1}{2}\times OP\times OQ=\frac{1}{2}\times4\times4=8

Answer

The normal to the curve at $A(2,-6)$ is $y=-x-4$ ; it meets the axes at $P(-4,0)$ and $Q(0,-4)$ , giving a triangle of area $8$ square units. Since $OP$ and $OQ$ both lie along the axes, which are already perpendicular, $\frac12\times OP\times OQ$ is the exact area, no extra trigonometry needed.

Q5[6 marks]

The curve $y=x^{3}-3x^{2}-9x+5$ has two turning points. Find the coordinates of both turning points, and use the second derivative to determine whether each is a maximum or a minimum.

Hence, state the range of values of $x$ for which the curve is decreasing.

Show worked solution

Turning points occur where $\frac{dy}{dx}=0$ . Differentiate and solve:

\frac{dy}{dx}=3x^{2}-6x-9=3(x-3)(x+1)=0 \;\Rightarrow\; x=3 \text{ or } x=-1

Substitute each $x$ -value back into the original equation to get the $y$ -coordinates:

y(-1)=(-1)^{3}-3(-1)^{2}-9(-1)+5=10 \qquad y(3)=(3)^{3}-3(3)^{2}-9(3)+5=-22

Use the second derivative to test the nature of each turning point:

\frac{d^{2}y}{dx^{2}}=6x-6 \;\Rightarrow\; \left.\frac{d^{2}y}{dx^{2}}\right|_{x=-1}=-12<0\;(\text{maximum}), \;\; \left.\frac{d^{2}y}{dx^{2}}\right|_{x=3}=12>0\;(\text{minimum})

The curve is decreasing wherever $\frac{dy}{dx}<0$ . Since $\frac{dy}{dx}=3(x-3)(x+1)$ is an upward parabola in $x$ with roots at $-1$ and $3$ , it is negative between the roots:

-1<x<3

Answer

There is a maximum at $(-1,10)$ and a minimum at $(3,-22)$ ; the curve decreases for $-1<x<3$ , exactly the interval between the two turning points, which makes sense, since the curve must fall from the maximum down to the minimum.

Q6[5 marks]

Air is pumped into a spherical balloon so that its volume increases at a constant rate of $32\pi$ cm $^3$ per second. Using $V=\frac{4}{3}\pi r^{3}$ for the volume and $S=4\pi r^{2}$ for the surface area of a sphere of radius $r$ , find the rate at which the surface area is increasing at the instant the radius is $4$ cm.

Show worked solution

First connect the rate we are given, $\frac{dV}{dt}$ , to the rate of change of the radius, $\frac{dr}{dt}$ , through the chain rule $\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}$ . Differentiate $V$ with respect to $r$ , then evaluate at $r=4$ :

\frac{dV}{dr}=4\pi r^{2} \;\Rightarrow\; \left.\frac{dV}{dr}\right|_{r=4}=4\pi(4)^{2}=64\pi

Substitute into the chain rule with the given $\frac{dV}{dt}=32\pi$ to find $\frac{dr}{dt}$ at this instant:

32\pi=64\pi\times\frac{dr}{dt} \;\Rightarrow\; \frac{dr}{dt}=\frac{1}{2}\text{ cm/s}

Now connect $\frac{dr}{dt}$ to the surface area through a second chain rule, $\frac{dS}{dt}=\frac{dS}{dr}\times\frac{dr}{dt}$ . Differentiate $S$ with respect to $r$ , evaluate at $r=4$ , then multiply:

\frac{dS}{dr}=8\pi r \;\Rightarrow\; \left.\frac{dS}{dr}\right|_{r=4}=8\pi(4)=32\pi

\frac{dS}{dt}=\frac{dS}{dr}\times\frac{dr}{dt}=32\pi\times\frac{1}{2}=16\pi

Answer

The surface area is increasing at $16\pi\ \text{cm}^{2}\text{/s}$ (about $50.3\ \text{cm}^{2}\text{/s}$ ) when $r=4$ cm. Both quantities are linked through the radius, so the trick is running the chain rule twice, once from $V$ to $r$ , once from $r$ to $S$ .

Q7[5 marks]

The curve $y=x\sqrt{2x+3}$ is defined for $x\ge-\frac{3}{2}$ . Using the product rule and the chain rule, find $\frac{dy}{dx}$ in its simplest single-fraction form, and hence find the coordinates of the turning point of the curve.

Show worked solution

Write the surd as a power and apply the product rule, $\frac{d}{dx}(uv)=u'v+uv'$ , with $u=x$ and $v=(2x+3)^{1/2}$ . Differentiating $v$ needs the chain rule:

\frac{dy}{dx}=(2x+3)^{\frac{1}{2}}+x\times\frac{1}{2}(2x+3)^{-\frac{1}{2}}\times2=(2x+3)^{\frac{1}{2}}+x(2x+3)^{-\frac{1}{2}}

Combine the two terms over the common factor $(2x+3)^{-1/2}$ to write this as a single fraction:

\frac{dy}{dx}=\frac{(2x+3)+x}{\sqrt{2x+3}}=\frac{3x+3}{\sqrt{2x+3}}=\frac{3(x+1)}{\sqrt{2x+3}}

At a turning point the numerator is zero (the denominator is never zero on the given domain, since $2x+3>0$ ):

3(x+1)=0 \;\Rightarrow\; x=-1

This lies in the domain $x\ge-\frac{3}{2}$ , so substitute back into the original equation for $y$ :

y=(-1)\sqrt{2(-1)+3}=(-1)\sqrt{1}=-1

Answer

$\frac{dy}{dx}=\frac{3(x+1)}{\sqrt{2x+3}}$ , and the turning point is at $(-1,-1)$ . Testing points either side, such as $x=-1.2$ and $x=-0.8$ , gives $y$ slightly above $-1$ in both cases, confirming it is a minimum.

Q8[6 marks]

The straight line $y=kx-9$ is a tangent to the curve $y=x^{2}-4x$ at the point $P$ . Find the two possible values of $k$ , and state the coordinates of $P$ in each case.

Show worked solution

At the point of tangency, the line and the curve must have the same gradient there. Differentiate the curve and set the result equal to $k$ :

\frac{dy}{dx}=2x-4=k

The line and the curve must also pass through the same point, so their $y$ -values agree at that $x$ . Substitute $k=2x-4$ into this condition:

x^{2}-4x=kx-9 \;\Rightarrow\; x^{2}-4x=(2x-4)x-9=2x^{2}-4x-9

Simplify this equation and solve for $x$ :

x^{2}-4x-2x^{2}+4x+9=0 \;\Rightarrow\; -x^{2}+9=0 \;\Rightarrow\; x^{2}=9 \;\Rightarrow\; x=3 \text{ or } x=-3

Find $k$ and the point $P$ for each value of $x$ , using $k=2x-4$ and $y=x^{2}-4x$ :

x=3: \;k=2(3)-4=2,\;y=9-12=-3 \;\Rightarrow\; P(3,-3)

x=-3: \;k=2(-3)-4=-10,\;y=9+12=21 \;\Rightarrow\; P(-3,21)

Answer

There are two tangent lines: $k=2$ touching at $P(3,-3)$ , and $k=-10$ touching at $P(-3,21)$ . Check either by substituting back, for instance at $x=3$ , the line gives $y=2(3)-9=-3$ , matching the curve.

Notice the shared shape of all three: turn the situation into a single equation you can differentiate, do the calculus carefully, then interpret the number in the language of the question, a rate, an approximate change, or a maximum area. That translation step is exactly what the harder marks reward.

Key method points

These examples pull the chapter's applications together. Keep the following points in mind for the longer, higher-mark questions.

Rate of change: chain the derivatives, $\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}$ , and evaluate $\frac{dy}{dx}$ at the given instant.
Small change: $\delta y\approx\frac{dy}{dx}\,\delta x$ , with $\frac{dy}{dx}$ taken at the starting value of $x$ .
Estimated value: add the approximate change to the known value, $y_{\text{new}}\approx y+\delta y$ .
Optimisation: build a function of one variable first, using the given constraint to eliminate the second variable.
Set $\frac{dA}{dx}=0$ to locate the stationary point, then confirm a maximum with $\frac{d^{2}A}{dx^{2}}<0$ .
Because marking is analytic, a correct relation and a clear derivative earn method marks even before the final number.

How a teacher helps

On these longer questions, the marks are won or lost in the setup, not the calculus. Students who can differentiate perfectly still stall when they cannot turn words into one clean equation.

In a one-to-one lesson our teacher works on exactly that translation, which variable to keep, which constraint to substitute, what the answer should mean, so the calculus becomes the easy last step. Because our teachers are experienced, you learn a repeatable way to read these problems.

Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I know which chain to build for a rate of change?

Write the rate you want and the rate you are given, then connect them through a shared variable. For $y$ and $x$ with $x$ changing in time, that link is $\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}$ .

Evaluate $\frac{dy}{dx}$ at the stated instant.

Why does the small-change formula use the gradient at the starting point?

The approximation $\delta y\approx\frac{dy}{dx}\,\delta x$ treats the curve as a straight line over the tiny interval. That line has the gradient at where you start, so you evaluate $\frac{dy}{dx}$ at the initial $x$ -value before multiplying by $\delta x$ .

In an optimisation problem, why must I get down to one variable?

You can only differentiate with respect to a single variable. Use the constraint, here the fixed $24$ m of fencing, to write one variable in terms of the other, so the quantity you are maximising becomes a function of just $x$ .

Do I always need the second derivative to confirm a maximum?

It is the cleanest confirmation and is expected for full marks. If $\frac{d^{2}A}{dx^{2}}<0$ the stationary point is a maximum; if it were positive, a minimum.

A sign check on $\frac{dA}{dx}$ either side also works.

Source:SRC-DSKP-EN

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

Differentiation, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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