Derivatives

• Calculate derivative using basic formulae- algebraic, trigonometric, inverse trigonometric, logarithmic and exponential.

• Calculate derivative using rules of derivatives such as - Addition/subtraction, product, quotient and chain rule.

• Apply differentiation to calculate tangent and normal, rate of change, maxima/minima and errors/approximation.

First Principle of Derivative

$f'(x) = \lim_{\Delta x \to 0}\frac{f(x+ \Delta x)-f(x)}{ \Delta x}$

Some Basic Formulae:

$\dfrac{d}{dx} \ c = 0$

$\dfrac{d}{dx} ( u \pm v \pm w \pm ....) = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \pm \dfrac{dw}{dx} \pm .....$

$\dfrac{d}{dx} ( cu ) = c \dfrac{du}{dx}$

$\dfrac{d}{dx} ( uv ) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$

$\dfrac{d}{dx} ( \dfrac {u}{v}) = \dfrac{v\frac{du}{dx} - u\dfrac{dv}{dx} }{v^{2}}$

Chain Rule :

$\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}$

List of Derivative Formulae:

Algebraic Group

$\dfrac{d}{dx} x^n = n x^{n-1}$

Trigonometric Group

$\dfrac{d}{dx} \sin x = \cos x$

$\dfrac{d}{dx} \cos x = -\sin x$

$\dfrac{d}{dx} \tan x = \sec^2 x$

$\dfrac{d}{dx} \cot x = - \csc^2 x$

$\dfrac{d}{dx} \sec x = \sec x \tan x$

$\dfrac{d}{dx} \csc x = - \csc x \cot x$

Logarithmic and Exponential Group

$\dfrac{d}{dx} e^x = e^x$

$\dfrac{d}{dx} a^x = a^x \log a$

$\dfrac{d}{dx} \log x = \dfrac{1}{x}$

Inverse Trigonometric Group

$\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}$

$\dfrac{d}{dx} \cos^{-1} x = \dfrac{-1}{\sqrt{1-x^2}}$

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

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

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

$\dfrac{d}{dx} \csc^{-1} x = \dfrac{-1}{x \sqrt{x^2 - 1}}$

Solved Example:

2-2-01

Equation of the line normal to function $f(x) = (x - 8)^{\frac{2}{3}} + 1$ at P(0, 5) is:

Solution:
\begin{align*} y &=\left( x-8\right) ^{\frac {2}{3}} + 1 \\ \dfrac {dy}{dx} &= \dfrac {2}{3}\left( x-8\right) ^{-\frac {1}{3}} \end{align*} \begin{equation*} \begin{split} \dfrac {dy}{dx}|_{x=0} &=\left( \dfrac {2}{3}\right) \left( -8\right) ^{-\frac {1}{3}}\\ & =\dfrac {2}{3}\left( \dfrac {1}{\sqrt [3] {-8}}\right) =\dfrac {2}{3}\left( \dfrac {1}{-2}\right) =\dfrac {-1}{3} \end{split} \end{equation*} $\mathrm{Slope\ of\ normal}=\dfrac {-1}{\left( -1/3\right) }=3$ \begin{align*} y-y_{1} &=m\left( x-x_{1}\right)\\ y-5 &=3\left( x-0\right)\\ y-5 &=3x\\ y &=3x+5 \end{align*}

Solved Example:

2-2-02

If $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$ then $\dfrac{dy}{dx}$ will be equal to:

Solution:
$\dfrac {dx}{d\theta }=a\left( 1+\cos \theta \right),\ \dfrac {dy}{d\theta }=a\left( 0+\sin \theta \right)$ $\dfrac {dy}{dx} =\dfrac {\left( \dfrac {dy}{d\theta }\right) }{\left( \dfrac {dx}{d\theta }\right) } =\dfrac {\sin \theta }{1+\cos \theta }\\ =\dfrac {2\sin \dfrac {\theta }{2}\cos \dfrac {\theta }{2}}{2\cos ^{2}\dfrac {\theta }{2}} = \tan\dfrac {\theta }{2}$

Solved Example:

2-2-03

For the following parametric function, what is the $\dfrac{d^2y}{dt^2}$ at t= 0? $x = 1 - t^2, \quad y= t- t^3$

Solution:
$y'=\dfrac {dy}{dx}$ Using chain rule, $y'=\dfrac {dy}{dt}\dfrac {dt}{dx}$ Since x is given in terms of parameter t, it is easier to find $\dfrac{dx}{dt}$ $y'=\dfrac {\dfrac {dy}{dt}}{\dfrac {dx}{dt}} =\dfrac {1-3t^{2}}{-2t}$ $y''=\dfrac {\dfrac {dy'}{dt}}{\dfrac {dx}{dt}} =\dfrac {\dfrac {-2t\left( 0-6t\right) -\left( 1-3t^{2}\right) \left( -2\right) }{\left( -2t\right) ^{2}}}{-2t}$ $y''=\dfrac {\dfrac {12t^{2}+2-6t^{2}}{4t^{2}}}{-2t} =\dfrac {\dfrac {2+6t^{2}}{4t^{2}}}{-2t} =\dfrac {-2-6t^{2}}{8t^{3}}$ $y''|_{t=0}=\dfrac {-2}{0}=\infty$

If $I_n= \dfrac{d^n}{dx^n}(x^n \log x),$ then: $I_n -n I_{n -1}= ?$