f ( x ) = 1 x f(x)=\dfrac {1}{\sqrt x} f ( x ) = x 1 , now to check that derivative exist or not , we know that to check the derivative of function exist or not , we apply formula -
f ′ ( x ) = l i m h → 0 f ( x + h ) − f ( x ) h f^{'}(x)= _{{lim\ h}{\to}0}\dfrac{f(x+h)-f(x)}{h} f ′ ( x ) = l im h → 0 h f ( x + h ) − f ( x )
, now l i m h → 0 _{limh\to0} l imh → 0 1 x + h − 1 x h \dfrac {\dfrac {1}{\sqrt {x+h}}-\dfrac{1}{\sqrt x}}{h} h x + h 1 − x 1
= l i m h → 0 =_{limh\to0} = l imh → 0 x − x + h x x + h h \dfrac {{ \dfrac {{\sqrt x}-{\sqrt{ x+h}}}{{\sqrt x}{\sqrt {x+h}}}}}{h} h x x + h x − x + h
= l i m h → 0 =_{limh\to0} = l imh → 0 x + x x + h − x + h x − ( x + h ) h x x + h ( x + x + h ) \dfrac{x+{\sqrt x}{\sqrt {x+h}}-{\sqrt{x+h}{\sqrt {x}}-(x+h)}}{h{\sqrt{x}{\sqrt {x+h}}}({\sqrt{x}+\sqrt{x+h)}}} h x x + h ( x + x + h ) x + x x + h − x + h x − ( x + h ) = 1 x 5 2 =\dfrac {1}{x^{\dfrac {5}{2}}} = x 2 5 1
hence we can clearly see that f'(0), of a function tends to infinity hence f'(0) does not exist .
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