NDA · Chapter

Differentiation

The concepts NDA needs, one connected idea at a time.

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NDA

NDA · Chapter

Differentiation

The concepts NDA needs, one connected idea at a time.

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concepts mastered

The derivative0/2
Rules0/4
Composite and inverse functions0/5
Other ways to differentiate0/4
Higher order derivatives0/3

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Up nextDerivative from first principles $f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$ :
The chain ruleStartIf $y = f(u)$ and $u = g(x)$ , then $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$ :

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How the ideas build on each other. Each one opens once you've mastered what it needs.

The derivative0 / 2

Derivative from first principles $f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$ :

the slope of the tangent

Start

differentiate the outside, keep the inside, multiply by the inside's derivative

the slope of the tangent

Start

Differentiability and continuityLeft and right derivatives must agree; differentiable implies continuous, but |x| is continuous with a corner at 0Locked

Rules0 / 4

Standard derivatives $\frac{d}{dx}x^n = nx^{n-1}$ , $\frac{d}{dx}\sin x = \cos x$ , $\frac{d}{dx}e^x = e^x$ , $\frac{d}{dx}\log x = \frac{1}{x}$ , $\frac{d}{dx}a^x = a^x\log a$ Locked
Sums and constant multiples $(u \pm v)' = u' \pm v'$
The product rule $(uv)' = uv' + vu'$ ,
The quotient rule $\left(\frac{u}{v}\right)' = \frac{vu' - uv'}{v^2}$ :

Composite and inverse functions0 / 5

The chain ruleIf $y = f(u)$ and $u = g(x)$ , then $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$ : differentiate the outside, keep the inside, multiply by the inside's derivativeStart
Chain rule with exponentials and logarithms $\dfrac{d}{dx}e^{u} = e^{u}u'$ ,
Derivative of an inverse functionIf $y = f(x)$ has inverse $x = g(y)$ , then $g'(y) = \dfrac{1}{f'(x)}$
Derivatives of inverse trigonometric functions $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1 - x^2}}$ ,
Simplifying by substitution firstPut $x = \tan\theta$ , $\sin\theta$ or $\cos\theta$ so that $\tan^{-1}\dfrac{2x}{1 - x^2}$

Other ways to differentiate0 / 4

Logarithmic differentiationTake $\log$ of both sides to turn products, quotients and powers like $x^{x}$ into sums, differentiate, then multiply back by $y$ Locked
Implicit differentiationDifferentiate every term in $x$ , writing $\dfrac{d}{dx}(y^n) = ny^{n-1}\dfrac{dy}{dx}$ ,
Parametric functionsIf $x = f(t)$ and $y = g(t)$ , then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ ,
Differentiating one function with respect to anotherThe derivative of $u$ with respect to $v$ is $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$

Higher order derivatives0 / 3

Second and higher derivativesDifferentiate $\dfrac{dy}{dx}$ again for $\dfrac{d^2y}{dx^2}$ ; for $e^{ax}$ , $\sin ax$ and similar, spot the $n$ th derivative's patternLocked
Second derivative of parametric and implicit functionsFor $x = f(t)$ , $y = g(t)$ : $\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\Big/\dfrac{dx}{dt}$ ,
Showing a relation between derivativesTo show $(1 - x^2)y_2 - xy_1 + m^2y = 0$ :

and

(ku)' = ku'

;

expand or split first

Locked

not

u'v'

Locked

denominator first in the numerator

Locked

\dfrac{d}{dx}a^{u} = a^{u}\log a\cdot u'

\dfrac{d}{dx}\log u = \dfrac{u'}{u}

;

simplify logs of products first

Locked

at the matching point: the slopes of a graph and its mirror in

y = x

are reciprocals

Locked

\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2}

\dfrac{d}{dx}\sec^{-1}x = \dfrac{1}{|x|\sqrt{x^2 - 1}}

;

with the chain rule for

\sin^{-1}u

Locked

becomes

2\theta = 2\tan^{-1}x

then differentiate the simple form

Locked

then collect the

\dfrac{dy}{dx}

terms and solve

Locked

the slope of the curve traced as

t

varies

Locked

not the ratio of second derivatives

Locked

find

y_1

clear roots or fractions, differentiate again, then substitute back

Locked