True or False.
If f(x)=1/√xf(x) = 1/√xf(x)=1/√x , then f′(0)f'(0)f′(0) does not exist.
f(x)=1xf(x)=\dfrac {1}{\sqrt x}f(x)=x1 , 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)=lim h→0f(x+h)−f(x)hf^{'}(x)= _{{lim\ h}{\to}0}\dfrac{f(x+h)-f(x)}{h}f′(x)=lim h→0hf(x+h)−f(x)
, now limh→0_{limh\to0}limh→0 1x+h−1xh\dfrac {\dfrac {1}{\sqrt {x+h}}-\dfrac{1}{\sqrt x}}{h}hx+h1−x1
=limh→0=_{limh\to0}=limh→0 x−x+hxx+hh\dfrac {{ \dfrac {{\sqrt x}-{\sqrt{ x+h}}}{{\sqrt x}{\sqrt {x+h}}}}}{h}hxx+hx−x+h
=limh→0=_{limh\to0}=limh→0 x+xx+h−x+hx−(x+h)hxx+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)}}}hxx+h(x+x+h)x+xx+h−x+hx−(x+h) =1x52=\dfrac {1}{x^{\dfrac {5}{2}}}=x251
hence we can clearly see that f'(0), of a function tends to infinity hence f'(0) does not exist .
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