Method of Differentiation Notes

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Method of DifferentiationIntroduction

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After the notes

Formula sheet

35 results from this chapter, grouped the way the notes teach them.

First principle

$f'(x)$ by first principle: $\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
Geometric meaning of $\dfrac{dy}{dx}$ at $x_0$: Slope of the tangent to $y=f(x)$ at $x=x_0$
Tangent to $y=f(x)$ at $(x_1,y_1)$: $y-y_1=\left.\dfrac{dy}{dx}\right|_{x_1}(x-x_1)$

Standard derivatives

$\dfrac{d}{dx}\,x^n$: $nx^{n-1}$
$\dfrac{d}{dx}\,e^x$: $e^x$
$\dfrac{d}{dx}\,a^x$: $a^x\ln a,\ \ a>0$
$\dfrac{d}{dx}\ln x$: $\dfrac1x$
$\dfrac{d}{dx}\sin x,\ \ \dfrac{d}{dx}\cos x$: $\cos x,\ \ -\sin x$
$\dfrac{d}{dx}\tan x,\ \ \dfrac{d}{dx}\cot x$: $\sec^2x,\ \ -\csc^2x$
$\dfrac{d}{dx}\sec x,\ \ \dfrac{d}{dx}\csc x$: $\sec x\tan x,\ \ -\csc x\cot x$
$\dfrac{d}{dx}\sin^{-1}x,\ \ \dfrac{d}{dx}\cos^{-1}x$: $\dfrac{1}{\sqrt{1-x^2}},\ \ -\dfrac{1}{\sqrt{1-x^2}}$
$\dfrac{d}{dx}\tan^{-1}x,\ \ \dfrac{d}{dx}\cot^{-1}x$: $\dfrac{1}{1+x^2},\ \ -\dfrac{1}{1+x^2}$
$\dfrac{d}{dx}\sec^{-1}x,\ \ \dfrac{d}{dx}\csc^{-1}x$: $\dfrac{1}{|x|\sqrt{x^2-1}},\ \ -\dfrac{1}{|x|\sqrt{x^2-1}}$

Rules of differentiation

Sum rule: $\dfrac{d}{dx}(f_1\pm f_2)=f_1'\pm f_2'$
Constant multiple: $\dfrac{d}{dx}\big(kf(x)\big)=k\,f'(x)$
Chain rule ( $y=f(u),\ u=g(x)$ ): $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$
Product rule: $\dfrac{d(uv)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
Quotient rule: $\dfrac{d}{dx}\left(\dfrac uv\right)=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
Product of three functions: $(fgh)'=f'gh+fg'h+fgh'$
$\dfrac{dy}{dx}\cdot\dfrac{dx}{dy}$: $1$ , when $\dfrac{dy}{dx}$ exists and $\dfrac{dx}{dy}\neq0$
Is $\dfrac{d^2y}{dx^2}\cdot\dfrac{d^2x}{dy^2}=1$ ?: Not always

Logarithmic differentiation

When to use it: Products or quotients of many functions, and $[f(x)]^{g(x)}$
$H=\dfrac{fg}{k\ell}$ ⇒ $\dfrac{H'}{H}$: $\dfrac{f'}{f}+\dfrac{g'}{g}-\dfrac{k'}{k}-\dfrac{\ell'}{\ell}$
$[f(x)]^{g(x)}$ rewritten: $e^{\,g(x)\ln f(x)}$
$\dfrac{d}{dx}[f(x)]^{g(x)}$: $[f(x)]^{g(x)}\left\{g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}\right\}$

Parametric & one function w.r.t. another

$x=f(t),\ y=g(t)$ ⇒ $\dfrac{dy}{dx}$: $\dfrac{dy/dt}{dx/dt}=\dfrac{g'(t)}{f'(t)}$
Derivative of $f(x)$ w.r.t. $g(x)$: $\dfrac{f'(x)}{g'(x)}$
Substitution for $\sqrt{a^2-x^2}$: $x=a\sin\theta$ or $a\cos\theta$
Substitution for $\sqrt{a^2+x^2}$: $x=a\tan\theta$ or $a\cot\theta$
Substitution for $\sqrt{\dfrac{a+x}{a-x}}$ or $\sqrt{\dfrac{a-x}{a+x}}$: $x=a\cos\theta$ or $a\cos2\theta$
Substitution for $\sqrt{(a-x)(x-b)}$: $x=a\cos^2\theta+b\sin^2\theta$

Implicit functions

Implicit function: $f(x,y)=0$ where $y$ cannot be written as $\phi(x)$ , e.g. $x^3+y^3=1$
Method: Differentiate every term w.r.t. $x$ , treating $y$ as a function of $x$ ; collect the $\frac{dy}{dx}$ terms
$\dfrac{d}{dx}f(y)$: $f'(y)\,\dfrac{dy}{dx}$
Direct formula for $f(x,y)=0$: $\dfrac{dy}{dx}=-\dfrac{\partial f/\partial x}{\partial f/\partial y}$ — differentiate keeping the other variable constant

Last look

Quick revision

The checks to run before you sit a question on functions.

1Simplify first — an algebraic or trig identity often turns the function into something you can differentiate in one line.
2Inside a chain, differentiate from the outside in and multiply each layer's derivative.
3A variable in both the base and the exponent ( $x^x$ , $(\sin x)^{\ln x}$ ) needs a log first: $y=e^{g\ln f}$ .
4For $y=u+v$ with each term a power of a function, take logs of $u$ and $v$ separately — never of the whole sum.
5Infinite expressions such as $\sqrt{\sin x+\sqrt{\sin x+\dots}}$ : write $y=\sqrt{\sin x+y}$ , then differentiate.
6If $f'(c)$ cannot be found from the rules (e.g. $x^{1/3}\sin x$ at $0$ ), go back to the first principle or check LHD and RHD.
7Product rule works only when both factors are differentiable at the point.
8For $\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)$ -type expressions, substitute $x=\tan\theta$ and watch the interval of $\theta$ — the answer can change sign piecewise.
9Parametric: find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ separately, then divide.
10Implicit: after differentiating, put the point in before solving for $\frac{dy}{dx}$ — it saves algebra.