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 9

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.

Practice this conceptself-check · 4 quick reps

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 9

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

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.

Don't power-rule an exponential

\frac{d}{dx}(a^x) = a^x\ln a

, NOT

x\,a^{x-1}

. The power rule

\frac{d}{dx}(x^n)=nx^{n-1}

applies only when the BASE is the variable. When the variable is in the EXPONENT, the derivative carries the base unchanged and picks up a

\ln a

factor. (And

\frac{d}{dx}e^x = e^x

, since

\ln e = 1

Derivative of $\ln x$ is $1/x$ , not $\ln x$ or $x$

\frac{d}{dx}(\ln x) = \dfrac{1}{x}

. It is neither

\ln x

(that's its own integral mistake) nor

x

. For a general base,

\frac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}

— the extra

\ln a

lives in the DENOMINATOR here, the opposite of where it sits for

a^x

Drill 4 more on standard derivatives to memorise

Concept 3 of 9

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

Practice this conceptself-check · 4 quick reps

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]

The product rule is not the product of derivatives

(uv)' \neq u'v'

. The correct rule is

(uv)' = u'v + uv'

— differentiate one factor at a time and add. For

x^2\sin x

, the answer is

2x\sin x + x^2\cos x

, not

(2x)(\cos x)

Quotient rule — order and sign in the numerator matter

\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}

, NOT

\dfrac{uv' - u'v}{v^2}

and NOT

\dfrac{u'v + uv'}{v^2}

. The derivative-of-the-top term comes first and the two terms are SUBTRACTED. Flipping the order negates the whole answer.

Drill 3 more on product and quotient rules

Concept 4 of 9

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)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4DifferentiationMODERATE

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

, what is dy/dx at

x=\pi

[Q98 · Apr · 2018]

Don't forget the derivative of the inner function

\frac{d}{dx}\sin(3x^2) = \cos(3x^2)\cdot 6x

, NOT just

\cos(3x^2)

. The chain rule multiplies by the inner derivative

6x

; leaving it out is the single most common slip in the chapter. Every layer contributes its own factor.

Drill 8 more on the chain rule (composite functions)

Concept 5 of 9

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

Practice this conceptself-check · 4 quick reps

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 9

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

Practice this conceptself-check · 4 quick reps

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 9

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.).

Standard inverse-trig collapses

\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x \qquad \sin^{-1}\dfrac{2x}{1+x^2} = 2\tan^{-1}x \qquad \cos^{-1}\dfrac{1-x^2}{1+x^2} = 2\tan^{-1}x \qquad \cos^{-1}(\sin x) = \dfrac{\pi}{2} - x \qquad \tan^{-1}\dfrac{a-b}{1+ab} = \tan^{-1}a - \tan^{-1}b

Worked example

Differentiate

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

Practice this conceptself-check · 4 quick reps

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 9

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)

Practice this conceptself-check · 4 quick reps

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

Concept 9 of 9

Trigonometric Simplification Toolkit

Intuition

Across calculus a 'hard' derivative is often just an unsimplified trig expression. Before differentiating, scan for these standard forms and collapse them first — a root becomes a plain sum, a quotient becomes a single tangent, a product becomes one angle.

Definition

Keep these collapses in reflex memory:

**Half-angle of $1\pm\cos x$ :** $1-\cos x = 2\sin^2\tfrac{x}{2}$ , $1+\cos x = 2\cos^2\tfrac{x}{2}$ ; so $\dfrac{1-\cos x}{1+\cos x}=\tan^2\tfrac{x}{2}$ , $\dfrac{1}{1-\cos x}=\tfrac12\csc^2\tfrac{x}{2}$ , $\dfrac{1}{1+\cos x}=\tfrac12\sec^2\tfrac{x}{2}$ .
Power-reduction (double angle): $1-\cos 2x = 2\sin^2 x$ , $1+\cos 2x = 2\cos^2 x$ , $\sin 2x = 2\sin x\cos x$ .
Perfect square under a root: $1\pm\sin 2x=(\sin x\pm\cos x)^2$ and $1\pm\sin\theta=\left(\cos\tfrac\theta2\pm\sin\tfrac\theta2\right)^2$ , so $\sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|$ — keep the modulus; its sign depends on the interval.
** $\sec\pm\tan$ :** $\sec x+\tan x=\dfrac{1+\sin x}{\cos x}=\tan\!\left(\tfrac\pi4+\tfrac x2\right)$ , $\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)$ .
Harmonic form: $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\alpha)$ , so its extreme values are $\pm\sqrt{a^2+b^2}$ .
**Weierstrass $t=\tan\tfrac{x}{2}$ :** $\sin x=\dfrac{2t}{1+t^2}$ , $\cos x=\dfrac{1-t^2}{1+t^2}$ — useful whenever a rational function of $\sin x,\cos x$ must be handled in one variable.

The collapses you reach for most

1+\cos x = 2\cos^2\tfrac{x}{2},\quad 1-\cos x = 2\sin^2\tfrac{x}{2},\quad \sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|

\tfrac{x}{2}half-angle — appears whenever you collapse $1\pm\cos x$
|\cdots|the root of a perfect square is a MODULUS; fix the sign on the given interval

Worked example

Simplify

\sqrt{1+\sin 2x}

for

0<x<\tfrac{\pi}{4}

, then differentiate it.

Practice this conceptself-check · 4 quick reps

$1\pm\cos x$ (half-angle) vs $1\pm\cos 2x$ (power-reduction)

Different collapses:

1-\cos x = 2\sin^2\tfrac{x}{2}

but

1-\cos 2x = 2\sin^2 x

. Read the angle inside the cosine before choosing the factor — the wrong one halves or doubles the argument.

The root of a perfect square is a MODULUS

\sqrt{(\sin x-\cos x)^2} = |\sin x-\cos x|

, not

\sin x-\cos x

. Resolve the sign on the given interval: on

(\tfrac\pi4,\tfrac\pi2)

it is

+(\sin x-\cos x)

; on

(0,\tfrac\pi4)

it is

-(\sin x-\cos x)

. Dropping the modulus flips the sign of the whole derivative.

$\sec\pm\tan$ — mind which way the half-angle shifts

\sec x+\tan x=\tan\!\left(\tfrac\pi4+\tfrac x2\right)

but

\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)

. Useful check:

(\sec x+\tan x)(\sec x-\tan x)=\sec^2x-\tan^2x=1

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 (8)

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
$\frac{du}{dv} = \frac{du/dx}{dv/dx}$
Simplify the inverse-trig first, then differentiate
Standard inverse-trig collapses
$\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x \qquad \sin^{-1}\dfrac{2x}{1+x^2} = 2\tan^{-1}x \qquad \cos^{-1}\dfrac{1-x^2}{1+x^2} = 2\tan^{-1}x \qquad \cos^{-1}(\sin x) = \dfrac{\pi}{2} - x \qquad \tan^{-1}\dfrac{a-b}{1+ab} = \tan^{-1}a - \tan^{-1}b$
Differentiating functional equations
Exponential functional equation
$f(x+y)=f(x)f(y) \;\Rightarrow\; f'(x) = f'(0)\,f(x)$
Trigonometric Simplification Toolkit
The collapses you reach for most
$1+\cos x = 2\cos^2\tfrac{x}{2},\quad 1-\cos x = 2\sin^2\tfrac{x}{2},\quad \sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|$

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 (10)

Degrees must be converted to radians first
→ Standard derivatives to memorise
Don't power-rule an exponential
→ Standard derivatives to memorise
Derivative of $\ln x$ is $1/x$ , not $\ln x$ or $x$
→ Standard derivatives to memorise
The product rule is not the product of derivatives
→ Product and quotient rules
Quotient rule — order and sign in the numerator matter
→ Product and quotient rules
Don't forget the derivative of the inner function
→ The chain rule (composite functions)
Don't quotient-rule the raw inverse-trig
→ Simplify the inverse-trig first, then differentiate
$1\pm\cos x$ (half-angle) vs $1\pm\cos 2x$ (power-reduction)
→ Trigonometric Simplification Toolkit
The root of a perfect square is a MODULUS
→ Trigonometric Simplification Toolkit
$\sec\pm\tan$ — mind which way the half-angle shifts
→ Trigonometric Simplification Toolkit

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