Differentiation Worked Examples (Medium)

Q: When do I use the product rule instead of just expanding?

For a simple product like (2x+1)(x2−3)(2x+1)(x^{2}-3)(2x+1)(x2−3) you can do either, and expanding is a good check. The product rule becomes essential when the factors are harder to expand, so it is worth practising even when expanding would also work.

Q: What is the most common quotient-rule mistake?

Writing the numerator as uv′−u′vuv'-u'vuv′−u′v instead of u′v−uv′u'v-uv'u′v−uv′. The order matters. A reliable memory aid is that the derivative of the top comes first: u′v−uv′u'v-uv'u′v−uv′, all over v2v^{2}v2.

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

Evaluate d2ydx2\frac{d^{2}y}{dx^{2}}dx2d2y at the turning point. If it is negative, the point is a maximum; if positive, a minimum. It is the fastest test whenever the second derivative is easy to compute and non-zero.

What these examples cover

These medium Differentiation examples step up from single terms to the two combination rules and to the shape of a curve. You will use the product rule $\frac{d}{dx}(uv)=u'v+uv'$ on a product of two brackets, the quotient rule $\frac{d}{dx}\!\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}}$ on a fraction, and the second derivative to decide whether a turning point is a maximum or a minimum.

The numbers stay small and clean so the method is never hidden behind messy arithmetic. Cover the solution, work the whole question on paper, then check line by line, and when you differ, pin down the exact step where it happened.

That is the line worth learning from.

Worked examples

Work through all three. The habit that ties them together is naming your parts before you start: decide what $u$ and $v$ are, write down $u'$ and $v'$ , and only then assemble the rule.

Q1[4 marks]

Differentiate $y=(2x+1)(x^{2}-3)$ with respect to $x$ , using the product rule.

Show worked solution

Name the two factors and differentiate each. Let $u=2x+1$ and $v=x^{2}-3$ , so $u'=2$ and $v'=2x$ .

The product rule gives $\frac{dy}{dx}=u'v+uv'$ :

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

Expand each bracket, then collect like terms:

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

Answer

$\frac{dy}{dx}=6x^{2}+2x-6$ . Check by expanding first: $y=2x^{3}+x^{2}-6x-3$ , and differentiating term by term gives $6x^{2}+2x-6$ , the two routes agree.

Q2[4 marks]

A curve is given by $y=\dfrac{3x}{x^{2}+1}$ . Find $\frac{dy}{dx}$ , and hence the gradient of the curve at $x=1$ .

Show worked solution

This is a quotient, so let $u=3x$ and $v=x^{2}+1$ , giving $u'=3$ and $v'=2x$ . Apply $\frac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}$ :

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

Expand the numerator carefully and simplify, the $v^{2}$ in the denominator is left as it is:

\frac{dy}{dx}=\frac{3x^{2}+3-6x^{2}}{(x^{2}+1)^{2}}=\frac{3-3x^{2}}{(x^{2}+1)^{2}}=\frac{3(1-x^{2})}{(x^{2}+1)^{2}}

Now substitute $x=1$ for the gradient there:

\left.\frac{dy}{dx}\right|_{x=1}=\frac{3(1-1)}{(1+1)^{2}}=\frac{0}{4}=0

Answer

$\frac{dy}{dx}=\dfrac{3(1-x^{2})}{(x^{2}+1)^{2}}$ , and the gradient at $x=1$ is $0$ . A zero gradient means the curve has a turning point at $x=1$ .

Q3[6 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

Turning points occur where the gradient is zero. Differentiate and factorise:

\frac{dy}{dx}=3x^{2}-6x=3x(x-2)

Set $\frac{dy}{dx}=0$ , so $3x(x-2)=0$ , giving $x=0$ or $x=2$ . Find the matching $y$ -values from the curve:

y(0)=0-0+4=4, \qquad y(2)=8-12+4=0

So the turning points are $(0,4)$ and $(2,0)$ . To classify them, use the second derivative:

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

Evaluate the second derivative at each turning point. A negative value gives a maximum, a positive value gives a minimum:

\left.\frac{d^{2}y}{dx^{2}}\right|_{x=0}=-6<0, \qquad \left.\frac{d^{2}y}{dx^{2}}\right|_{x=2}=6>0

Answer

$(0,4)$ is a maximum point and $(2,0)$ is a minimum point. The second-derivative test is quickest here because $\frac{d^{2}y}{dx^{2}}$ is easy to evaluate and never ambiguous when it is non-zero.

Q4[3 marks]

Differentiate $y=(3x-2)^{4}$ with respect to $x$ , using the chain rule.

Show worked solution

This is a bracket raised to a power, so treat the bracket as a single block $u$ . Let $u=3x-2$ , so $y=u^{4}$ and $u'=3$ .

The chain rule gives $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$ , so differentiate the outer power first, then multiply by $u'$ :

\frac{dy}{dx}=4u^{3}\times 3=12u^{3}

Substitute $u=3x-2$ back in:

\frac{dy}{dx}=12(3x-2)^{3}

Answer

$\frac{dy}{dx}=12(3x-2)^{3}$ . The power drops by one to $3$ , and every chain-rule derivative here carries the constant multiplier $3$ from differentiating the bracket.

Q5[4 marks]

Find the equation of the tangent to the curve $y=x^{2}-4x+5$ at the point where $x=1$ .

Show worked solution

Find the point on the curve, then the gradient there using $\frac{dy}{dx}$ , before building the tangent line. At $x=1$ :

y=(1)^{2}-4(1)+5=2, \qquad \frac{dy}{dx}=2x-4 \Rightarrow \left.\frac{dy}{dx}\right|_{x=1}=-2

Substitute the point $(1,2)$ and gradient $-2$ into $y-y_{1}=m(x-x_{1})$ :

y-2=-2(x-1) \implies y=-2x+4

Answer

The tangent is $y=-2x+4$ . Check by substituting $x=1$ : $y=-2(1)+4=2$ , matching the point on the curve.

Q6[4 marks]

Find the equation of the normal to the curve $y=x^{3}-2x$ at the point where $x=1$ .

Show worked solution

Find the point on the curve and the gradient of the tangent there using $\frac{dy}{dx}$ . At $x=1$ :

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

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: $m_{\text{normal}}=-\frac{1}{1}=-1$ . Substitute the point $(1,-1)$ and this gradient into $y-y_{1}=m(x-x_{1})$ :

y-(-1)=-1(x-1) \implies y=-x

Answer

The normal is $y=-x$ . Check by substituting $x=1$ : $y=-1$ , matching the point on the curve, and the product of the two gradients, $1\times(-1)=-1$ , confirms the lines are perpendicular.

Q7[3 marks]

Given $y=x^{3}$ , use differentiation to find the approximate change in $y$ when $x$ increases from $2$ to $2.01$ .

Show worked solution

A small change $\delta x$ in $x$ produces an approximate change in $y$ of $\delta y \approx \frac{dy}{dx}\times \delta x$ . Here $\delta x = 2.01-2=0.01$ .

Differentiate first:

\frac{dy}{dx}=3x^{2}

Evaluate the derivative at $x=2$ , then multiply by $\delta x$ :

\delta y \approx 3(2)^{2}\times 0.01 = 12\times 0.01=0.12

Answer

$\delta y \approx 0.12$ . Checking directly, $2.01^{3}-2^{3}=0.120601$ , so the approximation is accurate to two decimal places.

Q8[4 marks]

The radius of a circle is increasing at a constant rate of $0.5\text{ cm/s}$ . Find the rate of increase of the area of the circle at the instant when the radius is $4\text{ cm}$ .

Show worked solution

Connect the two rates through the chain rule $\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}$ . The area of a circle is $A=\pi r^{2}$ , so differentiate with respect to $r$ :

\frac{dA}{dr}=2\pi r

Substitute $r=4$ and the given rate $\frac{dr}{dt}=0.5$ :

\frac{dA}{dt}=2\pi(4)(0.5)=4\pi

Answer

$\frac{dA}{dt}=4\pi \approx 12.6\text{ cm}^{2}\text{/s}$ . The area grows fastest when the radius is largest, since $\frac{dA}{dr}=2\pi r$ increases with $r$ .

Across all three, the winning habit is the same: name $u$ and $v$ , write their derivatives before assembling any rule, and keep the denominator of a quotient untouched while you simplify the top. Do that and these questions become steady, method-mark-friendly work.

Key method points

These examples build the combination rules and the second-derivative test, the tools most medium Differentiation questions lean on. Keep the following points in mind.

Product rule: $\frac{d}{dx}(uv)=u'v+uv'$ . Name $u,v$ and write $u',v'$ before assembling.
Quotient rule: $\frac{d}{dx}\!\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}}$ . The order in the numerator matters, and the denominator is $v^{2}$ .
A zero gradient locates a turning point; substitute the $x$ -values back into the curve for the $y$ -coordinates.
Second-derivative test: $\frac{d^{2}y}{dx^{2}}<0$ at a point means a maximum, $\frac{d^{2}y}{dx^{2}}>0$ means a minimum.
Where you can, check a product-rule answer by expanding first and differentiating term by term.
Analytic marking rewards a correctly set-up rule, so write the $u'v+uv'$ or quotient line clearly even before you simplify.

How a teacher helps

The medium slips are predictable: the quotient numerator written in the wrong order, a product rule missing one of its two terms, or the second-derivative test applied to the wrong point. In a one-to-one lesson our teacher makes you name $u$ and $v$ out loud before touching the rule, which quietly removes most of these errors.

Because our teachers are experienced, you get someone who shows the reasoning, not just the mechanics. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation stays familiar whichever version you sit.

Get 1-to-1 help.

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Frequently asked questions

When do I use the product rule instead of just expanding?

For a simple product like $(2x+1)(x^{2}-3)$ you can do either, and expanding is a good check. The product rule becomes essential when the factors are harder to expand, so it is worth practising even when expanding would also work.

What is the most common quotient-rule mistake?

Writing the numerator as $uv'-u'v$ instead of $u'v-uv'$ . The order matters.

A reliable memory aid is that the derivative of the top comes first: $u'v-uv'$ , all over $v^{2}$ .

How does the second derivative tell me maximum from minimum?

Evaluate $\frac{d^{2}y}{dx^{2}}$ at the turning point. If it is negative, the point is a maximum; if positive, a minimum.

It is the fastest test whenever the second derivative is easy to compute and non-zero.

What if the second derivative is zero at the turning point?

Then the test is inconclusive and you fall back on checking the sign of $\frac{dy}{dx}$ just before and just after the point. In these examples the second derivative is non-zero, so the quick test settles it.

Source:SRC-DSKP-EN

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

Differentiation, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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