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

First Principle of Derivatives

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

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

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

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

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

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

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

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

Homework for the topic Derivatives

Solutions for the topic Derivatives

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*}

Correct Answer: B

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

Correct Answer: C

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 $\dfrac{dx}{dt}$ \[y'=\dfrac {\dfrac {dy}{dt}}{\dfrac {dx}{dt}} =\dfrac {1-3t^{2}}{-2t}\] \[y''=\dfrac {\dfrac {dy'}{dt}}{\dfrac {dx}{dt}} =\dfrac {\dfrac {-2t\left( 0-6t\right) -\left( 1-3t^{2}\right) \left( -2\right) }{\left( -2t\right) ^{2}}}{-2t}\] \[y''=\dfrac {\dfrac {12t^{2}+2-6t^{2}}{4t^{2}}}{-2t} =\dfrac {\dfrac {2+6t^{2}}{4t^{2}}}{-2t} =\dfrac {-2-6t^{2}}{8t^{3}}\] \[y''|_{t=0}=\dfrac {-2}{0}=\infty\]

Correct Answer: A

Solved Example:

2-2-04

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

Solution:
\begin{align*} I_{n} &=\dfrac {d^{n}}{dx^{n}}\left( x^{n}\log x\right)\\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ \dfrac {d}{dx}\left(x^{n}\log x\right)\right]\\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n}\dfrac {d}{dx}\left( \log x\right) +\log x\dfrac {d}{dx}\left(x^n\right)\right] \\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n}\left(\dfrac{1}{x}\right)+\log x\left(nx^{n-1}\right)\right] \\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n-1}+n\log x\left(x^{n-1}\right)\right] \\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n-1}\right]+ n\dfrac {d^{n-1}}{dx^{n-1}}\left[\log x\left(x^{n-1}\right)\right] \\ I_n &=\left( n -1\right) !+n I_{n -1}\\ I_n-nI_{n-1}&=\left( n-1\right) ! \end{align*}

Correct Answer: D