### Mathematics Assignments | PDF | 8 to 12

PDF Files of Mathematics Assignments From VIII Standard to XII Standard PDF of mathematics Assignments for the students from VIII standard to XII standard.These assignments are strictly according to the CBSE and DAV Board Final question Papers

# Differentiation For Classes 11 & 12

Differentiation formulas and basic concepts for classes 11 and 12 strictly according to the CBSE syllabus. Basic formulas of calculus

## DERIVATIVE BY FIRST PRINCIPAL

$f^{'}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
$f^{'}(x)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$

## DIFFERENTIATION OF SOME IMPORTANT  FUNCTIONS

$\frac{d}{dx}(x)=1$

$\frac{d}{dy}(x^{2})=2x$

$\frac{d}{dx}(x^{3})=3x^{2}$
$\frac{d}{dx}(x^{4})=4x^{3}$
$\frac{d}{dx}x^{n}=nx^{n-1}$
$\frac{d}{dx}(ax+b)=a$
$\frac{d(constant)}{dx}=\frac{d(C)}{dx}=0$

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

$\frac{d}{dx}logx=\frac{1}{x}$
$\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{-1}{x^{2}}$

$\frac{d}{dx}\left(\frac{1}{x^{2}}\right)=\frac{-2}{x^{3}}$

$\frac{d}{dx}\left(\frac{1}{x^{3}}\right)=\frac{-3}{x^{4}}$

$\frac{d}{dx}\left(\frac{1}{x^{n}}\right)=\frac{-n}{x^{n+1}}$

## DIFFERENTIATION OF SOME TRIGONOMETRIC FUNCTIONS

$\frac{d}{dx}sinx=cosx$
$\frac{d}{dx}cosx=-sinx$
$\frac{d}{dx}tanx=sec^{2}x$
$\frac{d}{dx}cotx=-cosec^{2}x$
$\frac{d}{dx}secx=secx\;tanx$
$\frac{d}{dx}cosecx=-cosecx\;cotx$

## PRODUCT RULE OF DIFFERENTIATION

(uv)' = u'v + uv'   OR

$\frac{d(uv)}{dx}=\left(\frac{du}{dx}\right)v+u\left(\frac{dv}{dx}\right)$

Example : Differentiate y = x2sinx, w.r.t. x
Solution

$\frac{dy}{dx}=\left(\frac{d}{dx}x^{2}\right)sinx+x^{2}\left(\frac{d}{dx}sinx\right)$

$\frac{dy}{dx}=2x\:sinx+x^{2}\:cosx$

## QUOTIENT RULE OF DIFFERENTIATION

$\left(\frac{u}{v}\right)^{'}=\frac{u'v-uv'}{v^{2}}$    (OR)

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\left(\frac{d}{dx}u\right)v-u\left(\frac{d}{dx}v\right)}{v^{2}}$

Example: Differentiate   $\frac{x^{2}}{sinx}$ , w. r. t. x
Solution:

$\frac{dy}{dx}=\frac{\left(\frac{d}{dx}x^{2}\right)sinx-x^{2}\left(\frac{d}{dx}sinx\right)}{(sinx)^{2}}$

$\frac{dy}{dx}=\frac{2xsinx-x^{2}cosx}{(sinx)^{2}}$

## Chain Rule of finding the differentiations

$\frac{d}{dx}\left(ax+b\right)^{3}=3(ax+b)^{3-1}\frac{d}{dx}(ax+b)=3a(ax+b)^{2}$
Example: Differentiate f(x) = (sin3x)4  with respect to x
Solution: f(x) = (sin3x)4

$f'(x)=4(sin3x)^{4-1}\left(\frac{d}{dx}sin3x\right)$
$f'(x)=4(sin3x)^{3}cos3x\left(\frac{d}{dx}3x\right)$

$f'(x)=12x(sin3x)^{3}cos3x$

## CONCEPT OF DIFFERENTIABILITY

A function is formally considered differentiable if its derivative exists at each point in its domain.
For a function to be differentiable it nust be continuous.

DEFINITION OF DIFFERENTIABILITY
f(x) is said to be differentiable at the point x = a if the derivative f ‘(a) exists at every point in its domain. It is given by

$f^{'}(x)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$
This formula is derived from Lagranges Mean Value Theorem.
Putting  x = a + h, as x → a, h → 0 so this can be written as

$f'(a)=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

CALCULATING DIFFERENTIABILITY
For checking the differentiability of a function we have to calculate Left Hand Differentiability (LHD) and Right Hand Differentiability (RHD)

It is batter to use basic formula of finding the differentiability as given below

$LHD=\displaystyle\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$

$RHD=\displaystyle\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}$

If LHD = RHD then the function is differentiable

Note: If a function is differentiable at any point, it is necessarily continuous at that point.

Example: Examine the function  $f(x)=\left\{\begin{matrix}1+x,&x\leq 2\\5-x,&x>2\\\end{matrix}\right.$   for differentiability at x = 2.
Solution:
At x < 2, f(x) = 1 + x

$Lf'(2)=\displaystyle\lim_{x\to 2^{-}}\frac{f(x)-f(2)}{x-2}$

$Lf'(2)=\displaystyle\lim_{x\to 2^{-}}\frac{(1+x)-(1+2)}{x-2}$

$Lf'(2)=\displaystyle\lim_{x\to 2^{-}}\frac{x-2}{x-2}=1$

At x > 2 f(x) = 5 - x

$Rf'(2)=\displaystyle\lim_{x\to 2^{+}}\frac{f(x)-f(2)}{x-2}$

$Rf'(2)=\displaystyle\lim_{x\to 2^{+}}\frac{(5-x)-(5-2)}{x-2}$

$Rf'(2)=\displaystyle\lim_{x\to 2^{+}}\frac{-x+2}{x-2}$

$Rf'(2)=\displaystyle\lim_{x\to 2^{+}}\frac{-(x-2)}{x-2}=-1$

⇒ At x = 2, Lf ' (2) ≠ Rf ' (2)

⇒ LHD  ≠ RHD

⇒ f(x) is not differentiable at x = 2

DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS
$If\:y=sin^{-1}x\:\:then\:\:\frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}$
$If\:y=cos^{-1}x\:\:then\:\:\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^{2}}}$
$If\:y=tan^{-1}x\:\:then\:\:\frac{dy}{dx}=\frac{1}{1+x^{2}}$
$If\:y=cot^{-1}x\:\:then\:\:\frac{dy}{dx}=\frac{-1}{1+x^{2}}$
$If\:y=sec^{-1}x\:\:then\:\:\frac{dy}{dx}=\frac{1}{|x|\sqrt{x^{2}-1}}$
$If\:y=cosec^{-1}x\:\:then\:\:\frac{dy}{dx}=\frac{-1}{|x|\sqrt{x^{2}-1}}$