Differentiate the following functions with respect to x from first pri

1.	Differentiate the following functions with respect to x from first principles :(i) √x
Answer» ANSWER: $\displaystyle \boxed{{<klux>F</klux>'(x)=\frac{1}{2\sqrt{x}}}}$ Step-by-step explanation: Given : f ( x ) = √ x We have to DIFF. w.r.t. x We have first PRINCIPLE of derivative . If y = f ( x ) Then : $\displaystyle{f'(x)= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} }$ = > f ( x + h ) = ( √ x + h ) $\displaystyle{f'(x)= \lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} }$ Now rationalize it we get : $\displaystyle{f'(x)= \lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \times\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} }\\\\\\\displaystyle{f'(x)= \lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} }\\\\\\\displaystyle{f'(x)= \lim_{h \to 0}\frac{1}{(\sqrt{x+h}+\sqrt{x})} }$ Putting h = 0 $\displaystyle{f'(x)=\frac{1}{(\sqrt{x+0}+\sqrt{x})}}\\\\\\\displaystyle{f'(x)=\frac{1}{\sqrt{x}+\sqrt{x}} }\\\\\\\displaystyle{f'(x)=\frac{1}{2\sqrt{x}}}$ Hence we get answer.

Differentiate the following functions with respect to x from first principles :(i) √x

Answer»

ANSWER:

$\displaystyle \boxed{{<klux>F</klux>'(x)=\frac{1}{2\sqrt{x}}}}$

Step-by-step explanation:

Given :

f ( x ) = √ x

We have to DIFF. w.r.t. x

We have first PRINCIPLE of derivative .

If y = f ( x )

Then :

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

= > f ( x + h ) = ( √ x + h )

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

Now rationalize it we get :

$\displaystyle{f'(x)= \lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \times\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} }\\\\\\\displaystyle{f'(x)= \lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} }\\\\\\\displaystyle{f'(x)= \lim_{h \to 0}\frac{1}{(\sqrt{x+h}+\sqrt{x})} }$

Putting h = 0

$\displaystyle{f'(x)=\frac{1}{(\sqrt{x+0}+\sqrt{x})}}\\\\\\\displaystyle{f'(x)=\frac{1}{\sqrt{x}+\sqrt{x}} }\\\\\\\displaystyle{f'(x)=\frac{1}{2\sqrt{x}}}$