NDA Maths · Differentiation

Core Techniques — Standard Derivatives, Rules, Chain & Logarithmic

The everyday toolkit: the derivative as a limit, the standard-derivative table, the product/quotient/chain rules, and logarithmic differentiation for variable exponents.

All Differentiation notes NDA Maths strategy

Why this matters

This subtopic carries the bulk of the chapter. Almost every question is 'recognise which tool applies' — a standard derivative, the chain rule, log-differentiation for a power tower, or a simplify-first move on an inverse-trig mess. Get these reflexes right and most of Differentiation becomes mechanical.

Concept 1 of 8

The derivative as a limit (first principles)

Intuition

The derivative is the limit of the slope of a chord as its two points slide together — the instantaneous rate of change. Every rule below is a shortcut for this one limit, so a question that writes the limit out is really just asking for the derivative.

Definition

$f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ — the slope of the tangent at $x$ . Equivalently $f'(a)=\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$ . Geometrically it is the slope of the tangent line; physically, a rate of change.

First-principles definition

f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

Worked example

Find

f'(x)

for

f(x)=x^2

from first principles.

$f'(x)=\lim_{h\to 0}\dfrac{(x+h)^2 - x^2}{h}=\lim_{h\to 0}\dfrac{2xh + h^2}{h}$ .
Cancel $h$ : $\lim_{h\to 0}(2x + h) = 2x$ .

Answer:

f'(x)=2x

Practice this conceptself-check · 4 quick reps

Try it yourself

g(x)=\sqrt{25-x^2}

, what does

\lim_{x\to 1}\dfrac{g(x)-g(1)}{x-1}

equal?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
First-principles definition of $f'(x)$ ?
2.
Geometric meaning of $f'(a)$ ?
3.
$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ is?
4.
From first principles, derivative of $x^2$ ?

From the bank · past-year question

Example 1DifferentiationMODERATE

l_1 = \dfrac{d}{dx}(e^{\sin x})

l_2 = \lim_{h \to 0}\dfrac{e^{\sin(x+h)} - e^{\sin x}}{h}

l_3 = \displaystyle\int e^{\sin x}\cos x\,dx

, then which one of the following is correct?

[Q73 · Sep · 2017]

Drill 9 more on the derivative as a limit (first principles)

Concept 2 of 8

Standard derivatives to memorise

Intuition

A fixed table of derivatives underlies everything. Know it cold — power, trig, exponential, logarithmic, and inverse-trig — so that the rules below just stitch these together. Most EASY marks are a single lookup from this table.

Definition

Memorise these; the rules (product, quotient, chain) combine them. Angles are in radians — a degree argument must be converted first.

Function f(x)	Derivative f′(x)
$x^n$	$n\,x^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\sec x$	$\sec x\tan x$
$e^x$	$e^x$
$a^x$	$a^x\ln a$ The $\ln a$ factor is the most-forgotten part of the table.
$\ln x$	$\dfrac{1}{x}$
$\log_a x$	$\dfrac{1}{x\ln a}$
$\sin^{-1} x$	$\dfrac{1}{\sqrt{1-x^2}}$
$\tan^{-1} x$	$\dfrac{1}{1+x^2}$

Radians only. The chain rule extends each of these to a composite argument.

Practice this conceptself-check · 4 quick reps

Try it yourself

What is the derivative of

\cosec(x^{\circ})

with respect to

x

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\frac{d}{dx}(a^x)$ ?
2.
$\frac{d}{dx}(\tan^{-1}x)$ ?
3.
$\frac{d}{dx}(\log_a x)$ ?
4.
Angle unit assumed by the trig derivatives?

From the bank · past-year question

Example 2DifferentiationMODERATE

What is the derivative of

\cosec(x^{\circ})

[Q94 · Apr · 2023]

Degrees must be converted to radians first

\frac{d}{dx}\sin(x^{\circ}) = \frac{\pi}{180}\cos(x^{\circ})

, NOT

\cos(x^{\circ})

. The standard table holds only for radian arguments; a degree symbol injects a

\pi/180

factor by the chain rule.

Drill 4 more on standard derivatives to memorise

Concept 3 of 8

Product and quotient rules

Intuition

Differentiating a product is not the product of derivatives. Use

(uv)' = u'v + uv'

for products and the quotient rule for ratios. Linearity handles sums: constants pull out, and

(u\pm v)' = u' \pm v'

Definition

Product: $(uv)' = u'v + uv'$ .
Quotient: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ .
Linearity: $(au \pm bv)' = au' \pm bv'$ .

Product and quotient rules

(uv)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Worked example

Differentiate

y = x^2 \sin x

Product rule with $u=x^2$ , $v=\sin x$ .
$y' = (2x)(\sin x) + (x^2)(\cos x)$ .

Answer:

y' = 2x\sin x + x^2\cos x

Practice this conceptself-check · 4 quick reps

Try it yourself

Differentiate

y = \dfrac{x}{1+x}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$(uv)'$ ?
2.
$(u/v)'$ ?
3.
$\frac{d}{dx}(x^2\sin x)$ ?
4.
Is $(uv)' = u'v'$ ?

From the bank · past-year question

Example 3DifferentiationEASY

Directions for the following three (03) items : Read the following information and answer the three items that follow : Let

f(x) = x^{2} + 2x - 5

and

g(x) = 5x + 30

h(x) = 5f(x) - xg(x)

, then what is the derivative of

h(x)

[Q93 · Sep · 2019]

Drill 3 more on product and quotient rules

Concept 4 of 8

The chain rule (composite functions)

Intuition

To differentiate a function of a function, differentiate the outer function (leaving the inner alone) and multiply by the derivative of the inner. Peel the layers from outside in — this single rule is the most-used tool in the chapter.

Definition

$\dfrac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$ . For nested layers, multiply each layer's derivative: $\dfrac{d}{dx}f(g(h(x))) = f'(g(h(x)))\,g'(h(x))\,h'(x)$ .

Chain rule

\frac{d}{dx}\,f(g(x)) = f'(g(x))\cdot g'(x)

Worked example

Differentiate

y = \sin(3x^2)

Outer $\sin$ → $\cos(3x^2)$ , inner $3x^2$ → $6x$ .
Multiply: $y' = \cos(3x^2)\cdot 6x$ .

Answer:

y' = 6x\cos(3x^2)

Practice this conceptself-check · 4 quick reps

Try it yourself

Differentiate

y = e^{\sin x}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\frac{d}{dx}f(g(x))$ ?
2.
$\frac{d}{dx}\sin(3x^2)$ ?
3.
$\frac{d}{dx}e^{\sin x}$ ?
4.
$\frac{d}{dx}(\ln(\cos x))$ ?

From the bank · past-year question

Example 4DifferentiationMODERATE

y=e^{x^2}\sin 2x

, what is dy/dx at

x=\pi

[Q98 · Apr · 2018]

Drill 8 more on the chain rule (composite functions)

Concept 5 of 8

Logarithmic differentiation

Intuition

When the variable is in the exponent (

x^x

f(x)^{g(x)}

) or you face a long product/quotient of powers, take

\ln

of both sides first. Logs convert powers to products and products to sums, after which you differentiate implicitly. It also cracks infinite power towers.

Definition

Take $\ln y$ , simplify with log laws, then differentiate (the left side gives $\frac{1}{y}\frac{dy}{dx}$ ):

Variable exponent: $y = f(x)^{g(x)} \Rightarrow \ln y = g(x)\ln f(x)$ .
Product of powers: $\ln y$ splits into a sum, each term differentiated alone.
Power tower: $y = (f(x))^{y} \Rightarrow \ln y = y\ln f(x)$ (the exponent is the whole $y$ again).

Logarithmic differentiation

y = f(x)^{g(x)} \;\Rightarrow\; \frac{1}{y}\frac{dy}{dx} = g'(x)\ln f(x) + g(x)\frac{f'(x)}{f(x)}

Worked example

Differentiate

y = x^x

Take $\ln$ : $\ln y = x\ln x$ .
Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx} = \ln x + 1$ .
Multiply by $y=x^x$ : $\dfrac{dy}{dx} = x^x(\ln x + 1)$ .

Answer:

\dfrac{dy}{dx} = x^x(1 + \ln x)

Practice this conceptself-check · 4 quick reps

Try it yourself

y = (\cos x)^{(\cos x)^{\cdots\infty}}

, find

\dfrac{dy}{dx}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
First step for $y = f(x)^{g(x)}$ ?
2.
$\frac{d}{dx}(x^x)$ ?
3.
Left side after $\ln y$ differentiated?
4.
Power tower $y=a^{a^{\cdots}}$ becomes?

From the bank · past-year question

Example 5DifferentiationMODERATE

y=\cos x\cdot\cos 4x\cdot\cos 8x

, then what is

\frac{1}{y}\frac{dy}{dx}

x=\frac{\pi}{4}

equal to?

[Q72 · Sep · 2021]

Drill 6 more on logarithmic differentiation

Concept 6 of 8

Derivative of one function with respect to another

Intuition

'Differentiate

u

with respect to

v

' is NOT

\frac{du}{dx}

. Treat both as functions of

x

, differentiate each w.r.t.

x

, and divide:

\frac{du}{dv} = \frac{du/dx}{dv/dx}

Definition

To find $\dfrac{du}{dv}$ where $u=u(x)$ and $v=v(x)$ : compute $\dfrac{du}{dx}$ and $\dfrac{dv}{dx}$ , then $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$ . It is the same idea as parametric differentiation, with $x$ as the hidden parameter.

Derivative of u w.r.t. v

\frac{du}{dv} = \frac{du/dx}{dv/dx}

Worked example

Find the derivative of

x^2

with respect to

x^3

$\dfrac{d(x^2)}{dx} = 2x$ , $\dfrac{d(x^3)}{dx} = 3x^2$ .
Divide: $\dfrac{d(x^2)}{d(x^3)} = \dfrac{2x}{3x^2} = \dfrac{2}{3x}$ .

Answer:

\dfrac{2}{3x}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the derivative of

\sin^2 x

with respect to

\cos^2 x

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\frac{du}{dv}$ in terms of $x$ -derivatives?
2.
Derivative of $x^2$ w.r.t. $x^3$ ?
3.
Derivative of $\sin^2 x$ w.r.t. $\cos^2 x$ ?
4.
Is 'derivative of $u$ w.r.t. $v$ ' the same as $du/dx$ ?

From the bank · past-year question

Example 6DifferentiationMODERATE

What is the derivative of

e^x

with respect to

x^e

[Q91 · Apr · 2021]

Drill 4 more on derivative of one function with respect to another

Concept 7 of 8

Simplify the inverse-trig first, then differentiate

Intuition

A messy inverse-trig expression almost always collapses before you differentiate. The standard move: substitute

x=\tan\theta

(or

x=\sin\theta

), recognise a double-angle or sum identity, and reduce the whole thing to a simple multiple of an angle. Differentiating the simplified form is trivial.

Definition

Common collapses (memorise the substitutions):

$\tan^{-1}\!\dfrac{2x}{1-x^2},\ \sin^{-1}\!\dfrac{2x}{1+x^2},\ \cos^{-1}\!\dfrac{1-x^2}{1+x^2}$ : put $x=\tan\theta\Rightarrow 2\theta = 2\tan^{-1}x$ .
$\cos^{-1}(\sin x) = \tfrac{\pi}{2}-x$ ; $\tan^{-1}\!\dfrac{a-b}{1+ab}=\tan^{-1}a-\tan^{-1}b$ .

Differentiate the collapsed form (often $\pm 1$ , $\pm 2/(1+x^2)$ , etc.).

Worked example

Differentiate

y = \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)

Put $x=\tan\theta$ : $\dfrac{2x}{1-x^2} = \tan 2\theta$ , so $y = 2\theta = 2\tan^{-1}x$ .
Differentiate: $\dfrac{dy}{dx} = \dfrac{2}{1+x^2}$ .

Answer:

\dfrac{dy}{dx} = \dfrac{2}{1+x^2}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the slope of

y = \cos^{-1}(\sin x)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Substitution for $\tan^{-1}\frac{2x}{1-x^2}$ ?
2.
$\frac{d}{dx}\tan^{-1}\frac{2x}{1-x^2}$ ?
3.
$\cos^{-1}(\sin x)$ simplifies to?
4.
Do you differentiate before or after simplifying?

From the bank · past-year question

Example 7DifferentiationMODERATE

y = \cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right)

, then

\dfrac{dy}{dx}

is equal to

[Q82 · Sep · 2017]

Don't quotient-rule the raw inverse-trig

Differentiating

\tan^{-1}\frac{2x}{1-x^2}

directly with the chain + quotient rule is slow and error-prone. The intended path is the

x=\tan\theta

collapse to

2\tan^{-1}x

first.

Drill 6 more on simplify the inverse-trig first, then differentiate

Concept 8 of 8

Differentiating functional equations

Intuition

When a function is defined by a rule like

f(x+y)=f(x)f(y)

rather than a formula, differentiate the relation (or use first principles) to extract

f'

. The exponential law

f(x+y)=f(x)f(y)

forces

f'(x)=f'(0)\,f(x)

Definition

For $f(x+y)=f(x)f(y)$ : from first principles $f'(x)=f(x)\lim_{h\to 0}\dfrac{f(h)-1}{h}=f'(0)\,f(x)$ . More generally, differentiate the given relation w.r.t. one variable and substitute convenient values (often $y=0$ ) to expose $f'$ .

Exponential functional equation

f(x+y)=f(x)f(y) \;\Rightarrow\; f'(x) = f'(0)\,f(x)

Worked example

f(x+y)=f(x)f(y)

for all

x,y

and

f'(0)=2

, express

f'(x)

From the relation, $f'(x) = f'(0)\,f(x)$ .
With $f'(0)=2$ : $f'(x) = 2f(x)$ .

Answer:

f'(x) = 2f(x)

(so

f(x)=e^{2x}

up to

f(0)=1

Practice this conceptself-check · 4 quick reps

Try it yourself

f(x+y)=f(x)f(y)

and

f(0)=1

, what is

f'(5)

in terms of

f'(0)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$f(x+y)=f(x)f(y) \Rightarrow f'(x)=?$
2.
What must $f(0)$ equal for $f(x+y)=f(x)f(y)$ (non-trivial)?
3.
Such an $f$ is which standard function?
4.
Handy substitution to extract $f'$ ?

From the bank · past-year question

Example 8DifferentiationMODERATE

Let

f(x+y)=f(x)f(y)

for all

x

and

y

. Then what is

f'(5)

equal to [where

f'(x)

is the derivative of

f(x)

[Q89 · Apr · 2017]

Drill 1 more on differentiating functional equations

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

The derivative as a limit (first principles)
First-principles definition
$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
Product and quotient rules
Product and quotient rules
$(uv)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
The chain rule (composite functions)
Chain rule
$\frac{d}{dx}\,f(g(x)) = f'(g(x))\cdot g'(x)$
Logarithmic differentiation
Logarithmic differentiation
$y = f(x)^{g(x)} \;\Rightarrow\; \frac{1}{y}\frac{dy}{dx} = g'(x)\ln f(x) + g(x)\frac{f'(x)}{f(x)}$
Derivative of one function with respect to another
Derivative of u w.r.t. v
$\frac{du}{dv} = \frac{du/dx}{dv/dx}$
Differentiating functional equations
Exponential functional equation
$f(x+y)=f(x)f(y) \;\Rightarrow\; f'(x) = f'(0)\,f(x)$

Reference tables (1)

Standard derivatives to memorise11 rows

Function f(x)	Derivative f′(x)
$x^n$	$n\,x^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\sec x$	$\sec x\tan x$
$e^x$	$e^x$
$a^x$	$a^x\ln a$ The $\ln a$ factor is the most-forgotten part of the table.
$\ln x$	$\dfrac{1}{x}$
$\log_a x$	$\dfrac{1}{x\ln a}$
$\sin^{-1} x$	$\dfrac{1}{\sqrt{1-x^2}}$
$\tan^{-1} x$	$\dfrac{1}{1+x^2}$

Radians only. The chain rule extends each of these to a composite argument.

Watch out for (2)

Degrees must be converted to radians first
→ Standard derivatives to memorise
Don't quotient-rule the raw inverse-trig
→ Simplify the inverse-trig first, then differentiate

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1DifferentiationMODERATE

Directions for the following three (03) items : Read the following information and answer the three items that follow : A curve

y = me^{mx}

where

m > 0

intersects

y

-axis at a point

P

How much angle does the tangent at

P

make with

y

-axis ?

[Q77 · Sep · 2019]

Example 2DifferentiationMODERATE

y = \frac{x\sqrt{x^2-16}}{2} - 8\ln\left|x+\sqrt{x^2-16}\right|

, then what is

\frac{dy}{dx}

equal to?

[Q77 · Sep · 2022]

Example 3DifferentiationMODERATE

u=e^{ax}\sin bx

and

v=e^{ax}\cos bx

, then what is

u\frac{du}{dx}+v\frac{dv}{dx}

equal to?

[Q91 · Sep · 2018]

Example 4DifferentiationEASY

If f(x) is an even function, then which is correct?

[Q97 · Apr · 2018]

Example 5DifferentiationMODERATE

y=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})(1+x^{16})

, then what is

\frac{dy}{dx}

x=0

equal to?

[Q71 · Sep · 2021]

Drill every past-year question on this subtopic

49 questions from the bank — paginated, with cart and Word-export support.

Drill the 49 questions

Drill every past-year question on this subtopic

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