CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

Differentiation Chapter - 5, Class 12Continuity @ Differentiability

Method of explaining continuity and differentiable, method of finding the derivative by First Principal. Differentiation of polynomial functions, logarithmic functions , exponential functions and trigonometric functions, product rule, quotient rule in differentiation

DIFFERENTIABILITY

f(x) is said to be differentiable if left hand differentiation is equal to the right hand differentiation.
$\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}$

DERIVATIVE BY FIRST PRINCIPAL

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

PRODUCT RULE OF DIFFERENTIATION

$\left ( uv \right )^{'}=u^{'}v+uv^{'}$

QUOTIENT RULE OF DIFFERENTIATION

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

DIFFERENTIATION OF TRIGONOMETRIC AND OTHER FUNCTIONS

Derivative of polynomial functions
$\frac{d}{dx}\left ( Constant \right )=0$
$\frac{d}{dx}x^{n}=nx^{n-1}$
$\frac{d}{dx}\left ( x^{5} \right )=5x^{4}$
$\frac{d}{dx}\left ( ax+b \right )=a$
$\frac{d}{dx}\left ( ax^{2}+bx \right )=2ax+b$
$\frac{d}{dx}\left ( ax^{5}+bx^{4}+cx^{3}+20 \right )=5ax^{4}+4bx^{3}+3cx^{2}$
Derivative of logarithmic functions
$\frac{d}{dx}logx=\frac{1}{x}$
$\frac{d}{dx}\left ( \frac{1}{x} \right )=-\frac{1}{x^{2}}$
$\frac{d}{dx}\left ( \frac{1}{x^{2}} \right )=-\frac{2}{x^{3}}$
$\frac{d}{dx}\left ( \frac{1}{x^{3}} \right )=-\frac{3}{x^{4}}$
Derivative of exponential functions
$\frac{d}{dx}e^{x}=e^{x}$
$\frac{d}{dx}e^{2x}=2e^{2x}$
$\frac{d}{dx}e^{5x}=5e^{5x}$
$\frac{d}{dx}e^{ax}=ae^{ax}$
$\frac{d}{dx}a^{x}=a^{x}\frac{d}{dx}x\: loga=a^{x}\: loga$
$\frac{d}{dx}2^{x}=2^{x}\frac{d}{dx}x\: log\: 2=2^{x}\: log\: 2$
Derivative of trigonometric functions
$\frac{d}{dx}sinx=cosx$
$\frac{d}{dx}cosx=-sinx$
$\frac{d}{dx}tanx=sec^{2}x$
$\frac{d}{dx}cotx=-cosec^{2}x$
$\frac{d}{dx}secx=secx\; tanx$
$\frac{d}{dx}cosecx=-cosecx\; cotx$

DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

$\frac{d}{dx}sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}$
$\frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^{2}}}$
$\frac{d}{dx}sec^{-1}x=\frac{1}{x\sqrt{x^{2}-1}}$
$\frac{d}{dx}cosec^{-1}x=\frac{-1}{x\sqrt{x^{2}-1}}$
$\frac{d}{dx}tan^{-1}x=\frac{1}{1+x^{2}}$
$\frac{d}{dx}cot^{-1}x=\frac{-1}{1+x^{2}}$

Method of writing the derivatives

First order derivatives can be written as $\frac{dy}{dx}=y'=y_{1}$Second order derivatives can be written as $\frac{d^{2}y}{dx^{2}}=y''=y_{2}$Third order derivatives can be written as$\frac{d^{3}y}{dx^{3}}=y'''=y_{3}$

LOGARITHMIC FUNCTIONS

Any function of type  y = $log_{b}\: x$  is called logarithmic function
If base b is replaced by 10 then it is called common logarithm.
If base b is replaced by 'e' then it is called natural logarithm .
Natural logarithm is denoted by lnx.
In place of lnx we simple write logx $log\: mn=log\;m+log\;n$$log\frac{m}{n}=logm-logn$$log\: m^{n}=n\: log\: m$

ROLL'S THEOREM

If f(x) is a continuous function in [a, b] and is differentiable in (a, b) and f(a) = f(b) then there exist some c ∊ (a, b) such that f ' (c) = 0

Algorithm for proving Roll's Theorem

a) Explain the continuity of the function f(x) in close interval [a, b]

b) Explain the differentiable of the function f(x) in open interval (a, b). If function is differentiable then find f '(x).

c) Check whether f(a) = f(b)

d) If all the above three conditions are satisfied then there exist some c ∊ (a, b) such that f ' (c) = 0.
Use this equation to find the value of c.

If  c ∊ (a, b) then Roll's Theorem verified.

NOTE:  f '(x) is called the slope of tangent and when f '(x) = 0 then slope is become parallel to the x-axis.
In the figure below  red lines shows the tangents are parallel to the x-axis at all the points where f '(x) = 0

MEAN VALUE THEOREM

If f(x) is a continuous function in [a,b] and is differentiable in (a,b) then there exist some c ∊ (a, b)  such that    $f'(c)=\frac{f(b)-f(a)}{b-a}$

Algorithm for proving Mean Value Theorem

a) Explain the continuity of the function f(x) in close interval [a, b]

b) Explain the differentiable of the function f(x) in open interval (a, b). If function is differentiable then find f '(x).

c) When both the above conditions are satisfied the use  formula given below to find the value of c. $f'(c)=\frac{f(b)-f(a)}{b-a}$
If  c ∊ (a, b) then Mean Value Theorem is verified.