Differentiation Worked Examples (Easy)

Q: Do I have to rewrite fractions and roots before differentiating?

It is by far the safest route. Writing 4x\dfrac{4}{x}x4 as 4x−14x^{-1}4x−1, or x\sqrt{x}x as x1/2x^{1/2}x1/2, lets the power rule apply directly, so you avoid the common slip of trying to differentiate a fraction as it stands.

Q: What exactly does the gradient at a point mean?

It is the value of dydx\frac{dy}{dx}dxdy at that xxx. Differentiate to get the gradient function, then substitute the xxx-value. A negative result means the curve is sloping downward there.

Q: What is the one step students forget in the chain rule?

The derivative of the bracket. For y=(3x−1)4y=(3x-1)^{4}y=(3x−1)4 you bring the power down and reduce it, then you must also multiply by 333, the derivative of 3x−13x-13x−1. Leaving out that factor is the usual error.

What these examples cover

These easy Differentiation examples build the four moves the whole chapter rests on: using the power rule $\frac{d}{dx}x^{n}=nx^{n-1}$ on a polynomial, rewriting a reciprocal or a root as a power before differentiating, applying the chain rule to a bracket raised to a power, and finding the equation of a tangent at a point. Each uses small, clean numbers so you can follow every line without a calculator getting in the way.

Cover the solution, attempt the question in full on paper, and only then check line by line against our working. Where your answer differs, find the exact step where the two solutions part company, that single line is usually where the real learning is.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, rewrite into powers, differentiate term by term, then substitute carefully, runs through every one.

Q1[3 marks]

Given $y=x^{3}-4x^{2}+5$ , find $\frac{dy}{dx}$ and the gradient of the curve at the point where $x=2$ .

Show worked solution

Differentiate term by term with the power rule $\frac{d}{dx}x^{n}=nx^{n-1}$ ; the constant $5$ differentiates to $0$ .

\frac{dy}{dx}=3x^{2}-8x

The gradient of a curve at a point is the value of $\frac{dy}{dx}$ there. Substitute $x=2$ :

\left.\frac{dy}{dx}\right|_{x=2}=3(2)^{2}-8(2)=12-16=-4

Answer

$\frac{dy}{dx}=3x^{2}-8x$ , and the gradient at $x=2$ is $-4$ . A negative gradient tells you the curve is sloping downward as it passes that point.

Q2[3 marks]

Differentiate $y=3x^{2}+\dfrac{4}{x}$ with respect to $x$ .

Show worked solution

First rewrite the reciprocal term as a power, so the power rule applies directly. Here $\dfrac{4}{x}=4x^{-1}$ .

y=3x^{2}+4x^{-1}

Now differentiate term by term, keeping the negative index:

\frac{dy}{dx}=6x+4(-1)x^{-2}=6x-4x^{-2}

Finally, write the negative power back as a fraction so the answer matches the form of the question:

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

Answer

$\frac{dy}{dx}=6x-\dfrac{4}{x^{2}}$ . Rewriting $\dfrac{4}{x}$ as $4x^{-1}$ first is what lets the power rule do the work, trying to differentiate a fraction as it stands is where marks slip away.

Q3[3 marks]

Given $y=(3x-1)^{4}$ , find $\frac{dy}{dx}$ .

Show worked solution

This is a function of a function, so use the chain rule. Let $u=3x-1$ , so that $y=u^{4}$ .

\frac{dy}{du}=4u^{3}, \qquad \frac{du}{dx}=3

Multiply the two derivatives, then replace $u$ with $3x-1$ :

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

Answer

$\frac{dy}{dx}=12(3x-1)^{3}$ . The shortcut is: bring the power down, reduce it by one, then multiply by the derivative of the bracket $(3)$ , never forget that last factor.

Q4[4 marks]

A curve has equation $y=x^{2}-3x+4$ . Find the equation of the tangent to the curve at the point where $x=1$ .

Show worked solution

First find the point of contact by substituting $x=1$ into the curve:

y=(1)^{2}-3(1)+4=1-3+4=2

So the tangent touches the curve at $(1,2)$ . Next find the gradient function and evaluate it at $x=1$ :

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

The gradient of the tangent is $-1$ . Use $y-y_{1}=m(x-x_{1})$ with $m=-1$ and $(x_{1},y_{1})=(1,2)$ :

y-2=-1(x-1) \;\Rightarrow\; y=-x+1+2 \;\Rightarrow\; y=-x+3

Answer

The tangent is $y=-x+3$ . Check the point lies on it: at $x=1$ , $y=-1+3=2$ , which matches $(1,2)$ .

Q5[3 marks]

Differentiate $y=6\sqrt{x}+3x$ with respect to $x$ .

Show worked solution

Rewrite the root as a power before differentiating: $\sqrt{x}=x^{1/2}$ , so $y=6x^{1/2}+3x$ .

y=6x^{1/2}+3x

Differentiate term by term with the power rule $\frac{d}{dx}x^{n}=nx^{n-1}$ :

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

Write the negative power back as a root so the answer matches the form of the question:

\frac{dy}{dx}=\frac{3}{\sqrt{x}}+3

Answer

$\frac{dy}{dx}=\frac{3}{\sqrt{x}}+3$ . Rewriting $\sqrt{x}$ as $x^{1/2}$ first is what lets the power rule apply, the same move as turning a reciprocal into a negative power.

Q6[3 marks]

A curve has equation $y=x^{2}-8x+10$ . Find the coordinates of the turning point of the curve.

Show worked solution

At a turning point the gradient is zero, so first find $\frac{dy}{dx}$ .

\frac{dy}{dx}=2x-8

Set the gradient to zero and solve for $x$ :

2x-8=0 \;\Rightarrow\; x=4

Substitute $x=4$ back into the original equation to find the corresponding $y$ -value:

y=(4)^{2}-8(4)+10=16-32+10=-6

Answer

The turning point is $(4,-6)$ . Finding where $\frac{dy}{dx}=0$ first, then substituting back into $y$ , is the routine for every turning-point question.

Q7[3 marks]

Given $y=x^{3}-5x^{2}+2x$ , find $\frac{d^{2}y}{dx^{2}}$ , and its value when $x=2$ .

Show worked solution

Differentiate once to find $\frac{dy}{dx}$ , using the power rule term by term.

\frac{dy}{dx}=3x^{2}-10x+2

Differentiate a second time to find $\frac{d^{2}y}{dx^{2}}$ :

\frac{d^{2}y}{dx^{2}}=6x-10

Substitute $x=2$ to find its value at that point:

\left.\frac{d^{2}y}{dx^{2}}\right|_{x=2}=6(2)-10=2

Answer

$\frac{d^{2}y}{dx^{2}}=6x-10$ , and its value at $x=2$ is $2$ . The second derivative is simply the first derivative differentiated again, one more application of the same power rule.

Q8[3 marks]

The side length of a square is increasing at a constant rate of $2\text{ cm/s}$ . Find the rate at which the area of the square is increasing at the instant its side is $5\text{ cm}$ .

Show worked solution

Let the side be $x$ cm and the area be $A=x^{2}$ . We are given $\frac{dx}{dt}=2$ , and we want $\frac{dA}{dt}$ .

\frac{dA}{dx}=2x

Connect the two rates with the chain rule:

\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dt}=2x\times 2=4x

Substitute $x=5$ :

\frac{dA}{dt}=4(5)=20

Answer

The area is increasing at $20\text{ cm}^{2}\text{/s}$ when the side is $5\text{ cm}$ . Connecting rates always goes through the chain rule $\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dt}$ .

Different as these four look, the routine underneath is the same: get every term into power form, differentiate carefully, and substitute one clean step at a time. That steadiness turns Differentiation into a dependable source of marks in both papers.

Key method points

These four examples rehearse the skills that open almost every Differentiation question in Add Math. Keep the following points in mind as you practise more.

Differentiate a polynomial term by term with the power rule $\frac{d}{dx}x^{n}=nx^{n-1}$ ; any constant differentiates to $0$ .
Rewrite reciprocals and roots as powers, $\dfrac{4}{x}=4x^{-1}$ , $\sqrt{x}=x^{1/2}$ , before differentiating.
The gradient of a curve at a point is the value of $\frac{dy}{dx}$ at that point.
For a bracket raised to a power, use the chain rule: bring the power down, reduce it by one, and multiply by the derivative of the bracket.
For a tangent, find the point first, then the gradient, then substitute into $y-y_{1}=m(x-x_{1})$ .
Because marking is analytic, a clear derivative line can still earn method marks even if the final arithmetic slips.

How a teacher helps

When a student drops a mark on questions like these, it is usually a small, fixable habit, a fraction left un-rewritten before differentiating, or the derivative of the bracket forgotten in a chain-rule step. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains the why behind each step, not just the what. 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

Do I have to rewrite fractions and roots before differentiating?

It is by far the safest route. Writing $\dfrac{4}{x}$ as $4x^{-1}$ , or $\sqrt{x}$ as $x^{1/2}$ , lets the power rule apply directly, so you avoid the common slip of trying to differentiate a fraction as it stands.

What exactly does the gradient at a point mean?

It is the value of $\frac{dy}{dx}$ at that $x$ . Differentiate to get the gradient function, then substitute the $x$ -value.

A negative result means the curve is sloping downward there.

What is the one step students forget in the chain rule?

The derivative of the bracket. For $y=(3x-1)^{4}$ you bring the power down and reduce it, then you must also multiply by $3$ , the derivative of $3x-1$ .

Leaving out that factor is the usual error.

How do I write the equation of a tangent?

Find the point on the curve, find the gradient by substituting into $\frac{dy}{dx}$ , then put both into $y-y_{1}=m(x-x_{1})$ . Simplify to $y=mx+c$ form.

Source:SRC-DSKP-EN

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

Differentiation, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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