Differentiation Ch-5 Class 12 | Continuity @ Differentiability

Differentiation Chapter - 5, Class 12
Continuity @ 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 =   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 \[log\: mn=log\;m+log\;n\]\[log\frac{m}{n}=logm-logn\]\[log\: m^{n}=n\: log\: m\]

DELETER FROM CBSE SYLLABUS

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. 

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