Calculus · Paper 1
Differentiation — Part 3
First Principles · The Rule · Standard Form · Higher Level · Tap NEXT to begin
Section 1 of 11
What's Coming
Part 1 gave us the Rule. Part 2 added Product, Quotient and Chain. Now we handle the three function families that keep appearing in exam questions: logs, exponentials, and inverse trig.
Good news: every one of them is just the Chain Rule + a standard derivative from the tables. No new rules — just two short patterns to memorise.
If Chain Rule from Part 2 isn't solid yet, go back and tighten it before continuing. Everything below depends on it.
Section 2 of 11
Logs — The Rule
From the tables: $\;y = \ln x \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{1}{x}$.
Now what if you have $\ln$ of something bigger, like $\ln(5x+3)$? Chain Rule.
(i) $y = \ln(5x+3)$
$u = 5x+3$ $\dfrac{du}{dx} = 5$
$y = \ln u$ $\dfrac{dy}{du} = \dfrac{1}{u} = \dfrac{1}{5x+3}$
$\dfrac{dy}{dx} = \dfrac{du}{dx} \cdot \dfrac{dy}{du} = 5 \cdot \dfrac{1}{5x+3}$
$\dfrac{dy}{dx} = \dfrac{5}{5x+3}$
The Log Rule — Must learn
Differ the bracket × differ the log
"Differ the log" means: log becomes 1 over the bracket.
$y = \ln\bigl(g(x)\bigr) \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{g'(x)}{g(x)}$
(ii) $y = \ln(8x+3)$
Bracket-diff: $8$ · Log becomes $\dfrac{1}{8x+3}$
$\dfrac{dy}{dx} = \dfrac{8}{8x+3}$
(iii) $y = \ln(16x+3)$
Bracket-diff: $16$
$\dfrac{dy}{dx} = \dfrac{16}{16x+3}$
(iv) $y = \ln(x^{2} + 5x + 1)$
Same rule — the bracket doesn't have to be linear.
Bracket-diff: $2x + 5$ · Log becomes $\dfrac{1}{x^{2}+5x+1}$
$\dfrac{dy}{dx} = \dfrac{2x+5}{x^{2}+5x+1}$
(v) $y = \ln(x^{3} + 5x)$
Bracket-diff: $3x^{2} + 5$
$\dfrac{dy}{dx} = \dfrac{3x^{2}+5}{x^{3}+5x}$
Section 3 of 11
Logs — Your Turn
You try
Differentiate $\;y = \ln(7x+2)$.
Bracket-diff over the bracket.
Bracket-diff: $7$
$\dfrac{dy}{dx} = \dfrac{7}{7x+2}$
$\dfrac{7}{7x+2}$
You try
Differentiate $\;y = \ln(4x^{2} + 3)$.
Bracket-diff: $8x$. Place over the bracket.
Bracket-diff: $8x$
$\dfrac{dy}{dx} = \dfrac{8x}{4x^{2}+3}$
$\dfrac{8x}{4x^{2}+3}$
You try
Differentiate $\;y = \ln(2x^{3} - x)$.
Differentiate inside the bracket first, then place over the bracket.
Bracket-diff: $6x^{2} - 1$
$\dfrac{dy}{dx} = \dfrac{6x^{2}-1}{2x^{3}-x}$
$\dfrac{6x^{2}-1}{2x^{3}-x}$
Section 4 of 11
Logs Combined with Product & Quotient
Logs often turn up inside a Product or Quotient. Identify the rule first ($x = 10$ trick if you're not sure), then go.
(i) $y = x^{6} \ln x$ — Product
$u = x^{6}$ $v = \ln x$
$\dfrac{du}{dx} = 6x^{5}$ $\dfrac{dv}{dx} = \dfrac{1}{x}$
$\dfrac{dy}{dx} = x^{6} \cdot \dfrac{1}{x} + \ln x \cdot 6x^{5}$
$\dfrac{dy}{dx} = x^{5} + 6x^{5} \ln x$
(ii) $y = \dfrac{\ln x}{x^{2}}$ — Quotient
$u = \ln x$ $v = x^{2}$
$\dfrac{du}{dx} = \dfrac{1}{x}$ $\dfrac{dv}{dx} = 2x$
$\dfrac{dy}{dx} = \dfrac{x^{2} \cdot \dfrac{1}{x} - 2x \ln x}{(x^{2})^{2}}$
$\dfrac{dy}{dx} = \dfrac{x - 2x \ln x}{x^{4}}$
$\dfrac{dy}{dx} = \dfrac{x(1 - 2 \ln x)}{x^{4}}$ (factor $x$ from top)
$\dfrac{dy}{dx} = \dfrac{1 - 2 \ln x}{x^{3}}$
(iii) $y = \dfrac{x^{2}}{\ln x}$ — Quotient
$u = x^{2}$ $v = \ln x$
$\dfrac{du}{dx} = 2x$ $\dfrac{dv}{dx} = \dfrac{1}{x}$
$\dfrac{dy}{dx} = \dfrac{(\ln x)(2x) - x^{2} \cdot \dfrac{1}{x}}{(\ln x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{2x \ln x - x}{(\ln x)^{2}}$
Notice $(\ln x)^{2}$ on the bottom — that's $\ln x$ squared, NOT $\ln(x^{2})$. These are different things:
Careful — these are different
$\ln(x^{n}) = n \ln x$ — log rule, lets you bring the power down
$(\ln x)^{n}$ — $\ln x$ multiplied by itself $n$ times. No log rule applies.
Section 5 of 11
Logs — Use Log Rules First
If the inside of the log is a fraction, a power, or a root, use log rules to split it up before differentiating. The differentiation becomes much easier.
Log rules — Must learn
$\ln(AB) = \ln A + \ln B$
$\ln\!\left(\dfrac{A}{B}\right) = \ln A - \ln B$
$\ln(A^{n}) = n \ln A$
(i) $y = \ln\!\left(\dfrac{3x+1}{5x+3}\right)$
Split with log rules first.
$y = \ln(3x+1) - \ln(5x+3)$
Now differentiate each piece using the log rule:
$\dfrac{dy}{dx} = \dfrac{3}{3x+1} - \dfrac{5}{5x+3}$
(ii) $y = \ln(9x+1)^{5}$
Bring the power down first.
$y = 5 \ln(9x+1)$
$\dfrac{dy}{dx} = 5 \cdot \dfrac{9}{9x+1}$
$\dfrac{dy}{dx} = \dfrac{45}{9x+1}$
(iii) $y = \ln\!\sqrt{\dfrac{3x+1}{5x+6}}$
Square root is the same as power $\dfrac{1}{2}$. Bring it down, then split.
$y = \ln\!\left(\dfrac{3x+1}{5x+6}\right)^{\!\frac{1}{2}}$
$y = \dfrac{1}{2} \ln\!\left(\dfrac{3x+1}{5x+6}\right)$
$y = \dfrac{1}{2}\bigl[\ln(3x+1) - \ln(5x+6)\bigr]$
$\dfrac{dy}{dx} = \dfrac{1}{2}\!\left[\dfrac{3}{3x+1} - \dfrac{5}{5x+6}\right]$
You try
Differentiate $\;y = \ln\!\left(\dfrac{2x+1}{x+3}\right)$.
Split using $\ln\!\left(\dfrac{A}{B}\right) = \ln A - \ln B$ before differentiating.
$y = \ln(2x+1) - \ln(x+3)$
$\dfrac{dy}{dx} = \dfrac{2}{2x+1} - \dfrac{1}{x+3}$
$\dfrac{2}{2x+1} - \dfrac{1}{x+3}$
You try
Differentiate $\;y = \ln(4x+1)^{3}$.
Bring the power down: $y = 3 \ln(4x+1)$.
$y = 3 \ln(4x+1)$
$\dfrac{dy}{dx} = 3 \cdot \dfrac{4}{4x+1}$
$\dfrac{dy}{dx} = \dfrac{12}{4x+1}$
$\dfrac{12}{4x+1}$
You try
Differentiate $\;y = x \ln x$.
Two pieces multiplied — Product Rule. $u = x$, $v = \ln x$.
Product Rule. $u = x$, $\;v = \ln x$.
$\dfrac{du}{dx} = 1$, $\;\dfrac{dv}{dx} = \dfrac{1}{x}$.
$\dfrac{dy}{dx} = x \cdot \dfrac{1}{x} + \ln x \cdot 1$
$\dfrac{dy}{dx} = 1 + \ln x$
$1 + \ln x$
Section 6 of 11
Exponentials — The Rule
From the tables: $\;y = e^{x} \;\Rightarrow\; \dfrac{dy}{dx} = e^{x}$. The only function whose derivative is itself.
For $e$ raised to something bigger — apply Chain Rule.
(i) $y = e^{7x+3}$
$u = 7x+3$ $\dfrac{du}{dx} = 7$
$y = e^{u}$ $\dfrac{dy}{du} = e^{u} = e^{7x+3}$
$\dfrac{dy}{dx} = 7 \cdot e^{7x+3}$
$\dfrac{dy}{dx} = 7 e^{7x+3}$
The $e^{x}$ Rule — Must learn
Differ the power × differ the $e$
"Differ the $e$" means: the $e$ stays the same.
$y = e^{g(x)} \;\;\Rightarrow\;\; \dfrac{dy}{dx} = g'(x)\, e^{g(x)}$
(ii) $y = e^{x^{2}+5x}$
Power-diff: $2x + 5$
$\dfrac{dy}{dx} = (2x+5)\, e^{x^{2}+5x}$
(iii) $y = e^{x^{3}+4x+1}$
Power-diff: $3x^{2} + 4$
$\dfrac{dy}{dx} = (3x^{2}+4)\, e^{x^{3}+4x+1}$
Section 7 of 11
Exponentials — Tricks and Combined
Two rewrite tricks that turn ugly exponentials into easy ones — and two Product/Quotient combos that come up a lot.
(i) $y = \dfrac{1}{e^{3x}}$ — rewrite first
Bring the $e^{3x}$ up using a negative power.
$y = e^{-3x}$
Power-diff: $-3$
$\dfrac{dy}{dx} = -3 e^{-3x}$
$\dfrac{dy}{dx} = -\dfrac{3}{e^{3x}}$
(ii) $y = \dfrac{e^{x^{2}}}{e^{7x}}$ — rewrite first
Same base divided — subtract the powers (index rule).
$y = e^{x^{2} - 7x}$
Power-diff: $2x - 7$
$\dfrac{dy}{dx} = (2x-7)\, e^{x^{2}-7x}$
Always try a rewrite first. Quotient Rule on $\dfrac{e^{x^{2}}}{e^{7x}}$ would work but it's far messier.
(iii) $y = x^{2} e^{x}$ — Product
$u = x^{2}$ $v = e^{x}$
$\dfrac{du}{dx} = 2x$ $\dfrac{dv}{dx} = e^{x}$
$\dfrac{dy}{dx} = x^{2} e^{x} + e^{x} \cdot 2x$
$\dfrac{dy}{dx} = x e^{x}(x + 2)$ (factor $x e^{x}$)
(iv) $y = \dfrac{e^{x}}{x}$ — Quotient
$u = e^{x}$ $v = x$
$\dfrac{du}{dx} = e^{x}$ $\dfrac{dv}{dx} = 1$
$\dfrac{dy}{dx} = \dfrac{x \cdot e^{x} - e^{x} \cdot 1}{x^{2}}$
$\dfrac{dy}{dx} = \dfrac{e^{x}(x - 1)}{x^{2}}$ (factor $e^{x}$ on top)
Section 8 of 11
Exponentials — Your Turn
You try
Differentiate $\;y = e^{5x+2}$.
Power-diff $\times$ $e$ stays the same.
Power-diff: $5$
$\dfrac{dy}{dx} = 5 e^{5x+2}$
$5 e^{5x+2}$
You try
Differentiate $\;y = e^{x^{2}+3x}$.
Differentiate the power: $2x + 3$. Multiply by $e$ (left as is).
Power-diff: $2x + 3$
$\dfrac{dy}{dx} = (2x+3)\, e^{x^{2}+3x}$
$(2x+3)\, e^{x^{2}+3x}$
You try
Differentiate $\;y = \dfrac{1}{e^{2x}}$.
Rewrite as $e^{-2x}$ first.
$y = e^{-2x}$
Power-diff: $-2$
$\dfrac{dy}{dx} = -2 e^{-2x}$
$\dfrac{dy}{dx} = -\dfrac{2}{e^{2x}}$
$-\dfrac{2}{e^{2x}}$
You try
Differentiate $\;y = x e^{x}$.
Product Rule. $u = x$, $v = e^{x}$.
Product Rule. $u = x$, $\;v = e^{x}$.
$\dfrac{du}{dx} = 1$, $\;\dfrac{dv}{dx} = e^{x}$.
$\dfrac{dy}{dx} = x e^{x} + e^{x} \cdot 1$
$\dfrac{dy}{dx} = e^{x}(x + 1)$
$e^{x}(x+1)$
Section 9 of 11
Inverse Trig — The Tables
What is $\sin^{-1}$? It's the inverse of sin — it gives you back the angle. If $\sin A = p$, then $\sin^{-1} p = A$.
Two derivatives come straight from the maths tables. Both have the variable in the form $\dfrac{x}{a}$ where $a$ is a constant.
From the tables — Must learn the format
$y = \sin^{-1}\!\left(\dfrac{x}{a}\right) \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{1}{\sqrt{a^{2} - x^{2}}}$
$y = \tan^{-1}\!\left(\dfrac{x}{a}\right) \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{a}{a^{2} + x^{2}}$
$a$ is a constant. Identify it, then plug straight in.
(i) $y = \sin^{-1}\!\left(\dfrac{x}{12}\right)$
Already in the form $\sin^{-1}\!\left(\dfrac{x}{a}\right)$ with $a = 12$. So $a^{2} = 144$.
$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{144 - x^{2}}}$
(ii) $y = \tan^{-1} x$
Write this as $\tan^{-1}\!\left(\dfrac{x}{1}\right)$ to spot the form. So $a = 1$, $a^{2} = 1$.
$\dfrac{dy}{dx} = \dfrac{1}{1 + x^{2}}$
(iii) $y = \sin^{-1}\!\left(\dfrac{x}{5}\right)$
$a = 5$, $\;a^{2} = 25$.
$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{25 - x^{2}}}$
(iv) $y = \tan^{-1}\!\left(\dfrac{x}{4}\right)$
$a = 4$, $\;a^{2} = 16$.
$\dfrac{dy}{dx} = \dfrac{4}{16 + x^{2}}$
Section 10 of 11
Inverse Trig — Chain and Product
When the inside is not in $\dfrac{x}{a}$ form — like $\tan^{-1}(10x)$ — the tables formula doesn't fit directly. Use Chain Rule with the standard forms:
Chain Rule versions
$y = \sin^{-1}\!\bigl(g(x)\bigr) \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{g'(x)}{\sqrt{1 - g(x)^{2}}}$
$y = \tan^{-1}\!\bigl(g(x)\bigr) \;\;\Rightarrow\;\; \dfrac{dy}{dx} = \dfrac{g'(x)}{1 + g(x)^{2}}$
Same pattern as logs: differ the inside $\times$ the standard derivative.
(i) $y = \tan^{-1}(10x)$
Inside-diff: $10$
$\dfrac{dy}{dx} = \dfrac{10}{1 + (10x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{10}{1 + 100x^{2}}$
(ii) $y = \sin^{-1}(7x)$
Inside-diff: $7$
$\dfrac{dy}{dx} = \dfrac{7}{\sqrt{1 - (7x)^{2}}}$
$\dfrac{dy}{dx} = \dfrac{7}{\sqrt{1 - 49x^{2}}}$
(iii) $y = x \tan^{-1} x$ — Product
$u = x$ $v = \tan^{-1} x$
$\dfrac{du}{dx} = 1$ $\dfrac{dv}{dx} = \dfrac{1}{1 + x^{2}}$
$\dfrac{dy}{dx} = x \cdot \dfrac{1}{1+x^{2}} + \tan^{-1} x \cdot 1$
$\dfrac{dy}{dx} = \dfrac{x}{1+x^{2}} + \tan^{-1} x$
Section 11 of 11
Inverse Trig — Your Turn
You try
Differentiate $\;y = \sin^{-1}\!\left(\dfrac{x}{9}\right)$.
In the form $\sin^{-1}\!\left(\dfrac{x}{a}\right)$ with $a = 9$. Use the tables formula.
$a = 9$, $\;a^{2} = 81$
$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{81 - x^{2}}}$
$\dfrac{1}{\sqrt{81 - x^{2}}}$
You try
Differentiate $\;y = \tan^{-1}\!\left(\dfrac{x}{3}\right)$.
$a = 3$, $a^{2} = 9$. Use $\dfrac{a}{a^{2}+x^{2}}$.
$a = 3$, $\;a^{2} = 9$
$\dfrac{dy}{dx} = \dfrac{3}{9 + x^{2}}$
$\dfrac{3}{9 + x^{2}}$
You try
Differentiate $\;y = \sin^{-1}(3x)$.
Inside is $3x$ — not in $\dfrac{x}{a}$ form. Use Chain Rule version.
Inside-diff: $3$
$\dfrac{dy}{dx} = \dfrac{3}{\sqrt{1 - (3x)^{2}}}$
$\dfrac{dy}{dx} = \dfrac{3}{\sqrt{1 - 9x^{2}}}$
$\dfrac{3}{\sqrt{1 - 9x^{2}}}$
You try
Differentiate $\;y = \tan^{-1}(2x)$.
Inside is $2x$. Chain Rule version: inside-diff over $1 + \text{(inside)}^{2}$.
Inside-diff: $2$
$\dfrac{dy}{dx} = \dfrac{2}{1 + (2x)^{2}}$
$\dfrac{dy}{dx} = \dfrac{2}{1 + 4x^{2}}$
$\dfrac{2}{1 + 4x^{2}}$
That's the differentiation toolkit.
Power · Product · Quotient · Chain · Logs · Exponentials · Inverse Trig. With these you can differentiate anything the exam will throw at you. Next up: applications — tangents, normals, max & min, rates of change.