Informative line

Derivative By First Principle

Learn definition of derivative as a function by first principle and find an expression for f?(x) by using definition of derivative. Practice differentiability of a function at a point & relationship between differentiability and continuity.

Derivative as a Function

\(f'{(a)}\) \(=\lim\limits_{h\to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)\)

Now let \('a'\) be a variable number \(x\), so that as \(x\) changes this expression changes with \(x\) and hence becomes a function of \(x\).

\(f'{(x)}\) \(=\lim\limits_{h\to 0}\;\left(\dfrac{f(x+h)-f(x)}{h}\right)\)

We say that this is derivative of f( i.e., f'), provided the limit exists at that \(x\).

  • It is the slope of tangent to the graph of \(f(x)\) at the point \((x,\,f(x))\).
  • The domain of \(f'\) may be smaller than the domain of \(f\).

(There may be some values of \(x\) for which \(f'\) may not be defined or limit does not exist).

Illustration Questions

If  \(f(x)=\dfrac{1}{2x+3}\),  find an expression for \(f'(x)\) by using definition of derivative.

A \(\dfrac{1}{x+7}\)

B \(\dfrac{-2}{(2x+3)^2}\)

C \(\dfrac{5x}{x+3}\)

D \(x^2\)

×

\(f'(x)\) \(=\lim\limits_{h\to 0}\;\left(\dfrac{f(x+h)-f(x)}{h}\right)\)

\(=\lim\limits_{h\to 0}\;\left(\dfrac{\dfrac{1}{2(x+h)+3}-\dfrac{1}{2x+3}}{h}\right)\)

\(=\lim\limits_{h\to 0}\;\left(\dfrac{2x+3-2x-2h-3}{h(2x+3)\,(2x+2h+3)}\right)\)

\(=\lim\limits_{h\to 0}\;\left(\dfrac{-2h}{h(2x+3)\,(2x+2h+3)}\right)\)

\(=\dfrac{-2}{(2x+3)^2}\)

If  \(f(x)=\dfrac{1}{2x+3}\),  find an expression for \(f'(x)\) by using definition of derivative.

A

\(\dfrac{1}{x+7}\)

.

B

\(\dfrac{-2}{(2x+3)^2}\)

C

\(\dfrac{5x}{x+3}\)

D

\(x^2\)

Option B is Correct

Other Notation used for Derivatives

We normally take \(y\) as the dependent variable and \(x\) as the independent variable.

\(y=f(x)\)

We use the notation:

derivatives of \(y\) with respect to \(x=f'(x)\)

Sometimes we use the notations,

\(f'(x)=y'=\underbrace{\dfrac{dy}{dx}=\dfrac{df}{dx}=\dfrac{df(x)}{dx}=Df(x)=D_xf(x)}_\text{All expression have the same meaning}\)

  • \(D\) and \(\dfrac{d}{dx}\) are called differentiation operators, because they indicate operation of differentiation.
  • The process of finding derivatives is called differentiation.
  • \(\dfrac{dy}{dx}\Bigg|_{x=a}\) = \(f'{(a)}\)is the other notation used.

Note that \(f'{(a)}\) means first take the derivative of the function with respect to \(x\)and then put \(x=a\) in the derivative expression. It is not the derivative of \(f(a)\), which will always be 0 because \(f(a)\) is a constant.

Illustration Questions

If  \(y=\dfrac{x-1}{x+1}\) , then find \(\dfrac{dy}{dx}\) using definition of derivative.

A \(\dfrac{4}{(x+1)^3}\)

B \(\dfrac{2}{(x+1)^2}\)

C \(\dfrac{2x}{x^5+7}\)

D \(4x^5\)

×

\(\dfrac{dy}{dx}=\) \(f'{(x)}\) \(=\lim\limits_{h\to 0}\;\left(\dfrac{f(x+h)-f(x)}{h}\right)\)

\(=\lim\limits_{h\to 0}\;\left(\dfrac{\dfrac{x+h-1}{x+h+1}-\dfrac{x-1}{x+1}}{h}\right)\)

\(=\lim\limits_{h\to 0}\;\dfrac{(x+1)(x+h-1)-(x-1)(x+h+1)}{h(x+1)(x+h+1)}\)

\(=\lim\limits_{h\to 0}\;\dfrac{x^2+xh-x+x+h-1-x^2-xh-x+x+h+1}{h(x+1)(x+h+1)}\)

\(=\lim\limits_{h\to 0}\;\dfrac{2h}{h(x+1)(x+h+1)}\)

\(=\dfrac{2}{(x+1)^2}\)

If  \(y=\dfrac{x-1}{x+1}\) , then find \(\dfrac{dy}{dx}\) using definition of derivative.

A

\(\dfrac{4}{(x+1)^3}\)

.

B

\(\dfrac{2}{(x+1)^2}\)

C

\(\dfrac{2x}{x^5+7}\)

D

\(4x^5\)

Option B is Correct

Differentiability of a Function at a Point

A function \('f'\) is said to be differentiable at \(x=a\) if \(f'{(a)}\) exists i.e the limit of definition of derivative exists.

i.e.

\(=\lim\limits_{h\to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)\) exists.

  • Among commonly used functions, \(f(x)=|x-a|\) is non differentiable at \(x=a\) because R.H.L. \(\neq\) L.H.L. for derivative limit.
  • In general \(g(x)=|f(x)|\) is non differentiable at all those values of \(x\) where \(f(x)=0\).

Illustration Questions

The value of \('x'\) for which \(f(x)=|x-1|\) is non differentiable is

A 2

B 5

C –7

D 1

×

\('f'\) is non differentiable at those values of \(x\) for which expression inside the modulus become 0.

\(\therefore\,x-1=0\)

\(\Rightarrow\,x=1\)

To verify consider,

right hand limit of derivative

\(=\lim\limits_{h\to 0^+}\;\left(\dfrac{f(1+h)-f(1)}{h}\right)\)

\(=\lim\limits_{h\to 0^+}\;\dfrac{|1+h-1|-0}{h}\)

\(=\lim\limits_{h\to 0^+}\;\dfrac{|h|}{h}\)  (\(|h|=h\) if \(h>0\))

\(=\lim\limits_{h\to 0^+}\;\dfrac{h}{h}\)

\(=1\)

Left hand limit of derivative, 

\(=\lim\limits_{h\to 0^-}\;\left(\dfrac{f(1+h)-f(1)}{h}\right)\)

\(=\lim\limits_{h\to 0^-}\;\dfrac{|1+h-1|-0}{h}\)

\(=\lim\limits_{h\to 0^-}\;\dfrac{|h|}{h}\)

\(=\lim\limits_{h\to 0}\;\dfrac{-h}{h}\)

\(=-1\)         (\(|h|=-h\) if \(h<0\))

\(\therefore\) Limit does not exist.

\(\Rightarrow\) Non differentiable at \(x=1\).

The value of \('x'\) for which \(f(x)=|x-1|\) is non differentiable is

A

2

.

B

5

C

–7

D

1

Option D is Correct

Relation Between Continuity and Differentiability at x=a

  • Continuity is a necessary but not sufficient condition for differentiable at a point \(x=a\).

i.e. if a function is continuous at \(x=a\), then it may or may not be differentiable at \(x=a\).

  • If a function is differentiable at \(x=a\), then it must be continuous at \(x=a\).
  • A broken graph will not allow a tangent to be drawn and hence no slope, so no \(f'\).

Illustration Questions

Given below is the graph of a certain function \(f\).The value of \('x'\) at which \(f\) fails to be differentiable is

A –2

B 2

C 5

D 7

×

Since \('f'\) is discontinuous at \(x=5\) , it is non differentiable at \(x=5\).

Given below is the graph of a certain function \(f\).The value of \('x'\) at which \(f\) fails to be differentiable is

image
A

–2

.

B

2

C

5

D

7

Option C is Correct

Differentiability in an Interval.

  • A function \('f'\) is differentiable in an open interval \((a,\,b)\) or \((a,\,\infty)\) or \((-\infty,\,a )\) if it is differentiable at each and every point in the interval.
  • If there is a non differentiable point in the interval \((a,\,b)\) then the function is said to be non differentiable in the entire interval.

\(f(x)=|x|\) is non differentiable at \(x=0\), in general \(|g(x)|\) will be non differentiable at all the roots of \(g(x)=0\).

Illustration Questions

Which of the following functions is differentiable in the interval \((-1,\,8)\) ?

A \(f(x)=|x-7|\)

B \(f(x)=|x|\)

C \(f(x)=|x-9|\)

D \(f(x)=|x-3|\)

×

For option (A),

\(f(x)=|x-7|\)  is non differentiable at \(x=7\)

\(\Rightarrow\,\) non differentiable in \((-1,\,8)\)

For option (B),

 \(f(x)=|x|\) is non differentiable at \(x=0\)

\(\Rightarrow\,\) non differentiable in \((-1,\,8)\)

For option (C),

 \(f(x)=|x-9|\) is differentiable at \(x=9\)

\(\Rightarrow\,\) differentiable in \((-1,\,8)\)

For option (D),

 \(f(x)=|x-3|\) is non differentiable at \(x=3\)

\(\Rightarrow\,\) non differentiable in \((-1,\,8)\)

Which of the following functions is differentiable in the interval \((-1,\,8)\) ?

A

\(f(x)=|x-7|\)

.

B

\(f(x)=|x|\)

C

\(f(x)=|x-9|\)

D

\(f(x)=|x-3|\)

Option C is Correct

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