Differentiation Techniques | A-Maths Study Guide

Secondary 4 Additional Mathematics · Study guide

Connect tangent gradients to rates of change, choose a differentiation rule and keep every domain restriction.

Concept → worked examples → practice20 min read · practice linked below

A derivative is a local rate of change

The gradient of a straight line is constant. For a curve, the gradient can change from point to point. The derivative $f'(x)=\dfrac{dy}{dx}$ gives the gradient of the tangent to $y=f(x)$ at each input where a finite derivative exists.

A secant joins two points. Its gradient is $\dfrac{f(x+h)-f(x)}h$ , where $h\ne0$ . As the second point approaches the first, this gradient approaches the tangent gradient. For $f(x)=x^2$ , the secant gradient is $2x+h$ , approaching $2x$ . This explains what differentiation measures; the rules below let us calculate it efficiently.

If $x$ is measured in seconds and $y$ in metres, $dy/dx$ has units metres per second. Keep the original function's domain: an algebraic simplification does not restore an excluded input.

A secant uses two distinct points. A tangent captures the limiting gradient at one point.

Standard derivatives and their conditions

Derivatives used throughout A Maths
Function	Derivative	Condition
$x^n$	$nx^{n-1}$	Rational $n$ ; use where the real function is differentiable. Negative powers exclude zero; fractional powers may restrict the domain.
$c$	$0$	A constant has no change.
$e^x$	$e^x$	Every real input.
$\ln x$	$1/x$	$x>0$ .
$\sin x$	$\cos x$	Angles in radians.
$\cos x$	$-\sin x$	Angles in radians.
$\tan x$	$\sec^2x$	Radians; $\cos x\ne0$ .

Differentiate sums term by term and retain constant multipliers. The familiar trigonometric derivatives assume radians: if the input were a numerical angle in degrees, converting to radians would introduce a factor $\pi/180$ .

Example 1: rewrite roots and reciprocals as powers

Differentiate $y=3x^4+2\sqrt{x}-5/x^3+7$ .

The original domain is $x>0$ , because of both the square root and the denominator. Write $y=3x^4+2x^{1/2}-5x^{-3}+7$ , then apply the power rule to each term:

\frac{dy}{dx}=12x^3+x^{-1/2}+15x^{-4}=12x^3+\frac1{\sqrt{x}}+\frac{15}{x^4},\qquad x>0.

The constant 7 contributes zero. The reciprocal term becomes positive because $(-5)(-3)=15$ .

Chain rule: account for the inner change

If $y=F(u)$ and $u=g(x)$ , then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}$ , wherever both derivatives exist. A small change in $x$ first changes $u$ , then changes $y$ ; the two local scale factors multiply. Thus

\frac{d}{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x),\qquad \frac{d}{dx}e^{g(x)}=e^{g(x)}g'(x).

Example 2: a square root and its endpoint

Find the derivative of $y=\sqrt{3x-1}$ . The function exists for $x\ge\tfrac13$ .

\frac{dy}{dx}=\frac12(3x-1)^{-1/2}\cdot3=\frac{3}{2\sqrt{3x-1}},\qquad x>\tfrac13.

The derivative's domain is smaller: at $x=\tfrac13$ the denominator tends to zero, so there is no finite tangent gradient. At $x=\tfrac23$ , the gradient is $3/2$ .

Example 3: a chain with two inner layers

Differentiate $y=3\tan^2(x/2)$ , using radians. The function excludes $x=(2k+1)\pi$ , for any integer $k$ .

Differentiate the square, then the tangent, then $x/2$ :

\frac{dy}{dx}=6\tan(x/2)\cdot\sec^2(x/2)\cdot\frac12=3\tan(x/2)\sec^2(x/2).

At $x=0$ , the derivative is 0. The excluded inputs stay excluded even though a calculator may display a very large finite value near them.

Product and quotient rules

When both factors vary, changing either factor contributes to the change of their product:

\frac{d}{dx}(uv)=u^{\prime}v+uv^{\prime}.

For a quotient, write $u/v=u\,v^{-1}$ , then apply the product and chain rules. This gives

\frac{d}{dx}\left(\frac uv\right)=\frac{u'v-uv'}{v^2},\qquad v\ne0.

Keep the order in the numerator. Simplify first if expansion or division makes the expression easier, but carry all original restrictions forward.

Example 4: an exponential times a trigonometric function

For $y=e^{2x}\cos3x$ , find the gradient when $x=\pi/6$ .

y'=2e^{2x}\cos3x-3e^{2x}\sin3x=e^{2x}(2\cos3x-3\sin3x).

Both inner multipliers, 2 and 3, matter. Substitution gives

y'(\pi/6)=e^{\pi/3}(2\cdot0-3\cdot1)=-3e^{\pi/3}.

The tangent is descending there. The exponential itself is always positive; it never creates a zero of the derivative.

Example 5: a quotient with a forbidden input

Differentiate $f(x)=e^{3x}/(x-1)$ , where $x\ne1$ .

f'(x)=\frac{3e^{3x}(x-1)-e^{3x}}{(x-1)^2}=\frac{e^{3x}(3x-4)}{(x-1)^2},\qquad x\ne1.

The numerator is zero at $x=4/3$ . The value $x=1$ is neither a point on the curve nor a stationary point. Squaring the denominator does not remove its restriction.

Logarithms: simplify only within the domain

For a positive differentiable function $g(x)$ , $\dfrac{d}{dx}\ln g(x)=g'(x)/g(x)$ . Log laws often shorten the work, but $\ln A+\ln B$ requires $A>0$ and $B>0$ . A positive product alone does not guarantee both factors are positive.

Example 6: use log laws without enlarging the domain

Differentiate $y=\ln[e^x(x^2-4)]$ . Since $e^x>0$ , the log argument is positive exactly when $x<-2$ or $x>2$ .

On that domain, $y=x+\ln(x^2-4)$ , so

y'=1+\frac{2x}{x^2-4}=\frac{x^2+2x-4}{x^2-4},\qquad x<-2\text{ or }x>2.

Do not replace $\ln(x^2-4)$ by $\ln(x-2)+\ln(x+2)$ on the negative branch: those two logarithms would not be real. The rational derivative expression alone cannot tell you the original logarithm's full restriction.

Second derivatives and derivative relations

The second derivative $y''=d^2y/dx^2$ is the derivative of the gradient function. It describes how the gradient changes. It is not $(y')^2$ . For a proposed relation involving $y,y',y''$ , calculate the derivatives, substitute, then compare like terms.

Example 7: check a relation between a function and its derivatives

For $y=e^{2x}+3e^{-x}$ , verify that $y''=y'+2y$ .

y'=2e^{2x}-3e^{-x},\qquad y''=4e^{2x}+3e^{-x}.

y'+2y=(2+2)e^{2x}+(-3+6)e^{-x}=4e^{2x}+3e^{-x}=y''.

The identity holds for every real $x$ . If unknown constants replaced the coefficients 1 and 2 in the relation, equating the coefficients of $e^{2x}$ and $e^{-x}$ would give two equations.

Example 8: simplify first, but keep the hole

Differentiate $y=(x^2-1)/(x-1)$ , with $x\ne1$ .

Factor the numerator: $y=(x-1)(x+1)/(x-1)=x+1$ for $x\ne1$ . Therefore $y'=1$ for $x\ne1$ .

The graph is the line $y=x+1$ with the point $(1,2)$ missing. Its derivative is not defined at $x=1$ , because the original function has no value there. Cancelling a factor simplifies a formula, not its stated domain.

Explore a moving tangent

For $y=x^2$ , the tangent at $x=a$ is $y=2ax-a^2$ . Its slope is $2a$ . Move the point across the origin and compare descending, horizontal and ascending tangents.

Point input a:1

The solid curve is y = x². The dashed line is its tangent at the marked point.

At a = 1, the point is (1, 1) and the tangent gradient is 2. Tangent: y = 2x − 1.

Common mistakes and useful checks

Missing the inner derivative: $d(\sin3x)/dx=3\cos3x$ .
Multiplying derivatives for a product: use $u\,v$ → $u^{\prime}v+uv^{\prime}$ , not $u^{\prime}v^{\prime}$ .
Reversing the quotient numerator: top derivative times bottom comes first.
Using degree mode: calculus trigonometry here uses radians.
Dropping the domain: square roots, logs, denominators and cancelled factors still restrict inputs.
Squaring the first derivative: differentiate it again to obtain the second.

Continue with topic practice

Use the Secondary 4 differentiation techniques practice bank for exam questions and their worked solutions. Attempt a question on paper before comparing methods. This guide supports Secondary 4 learning and revision of the full Secondary Additional Mathematics course.