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).

#### 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.

#### 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

# 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 $$f'$$.

#### 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 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 $$g(x)=0$$.

#### 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

# 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 $$f(x)=0$$.

#### 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