Question

Differentiate the following functions with respect to x from first principles :

(i) √x

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Answer 1

Answer:

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

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.