Using the limit definition, how do you differentiate #f(x)=x^(-1/2)#?

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Answer 1 · 2018-06-30T23:32:01

Given #f(x)=x^{-1/2}=1/\sqrtx#

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

#=\lim_{h\to 0}\frac{1/\sqrt{x+h}-1/\sqrtx}{h}#

#=\lim_{h\to 0}\frac{1/{\sqrtx\sqrt{1+h/x}}-1/\sqrtx}{h}#

#=1/\sqrtx\lim_{h\to 0}\frac{1/(1+h/x)^{1/2}-1}{h}#

#=1/\sqrtx\lim_{h\to 0}\frac{(1+h/x)^{-1/2}-1}{h}#

#=1/\sqrtx\lim_{h\to 0}\frac{(1-1/2(h/x)+\frac{(-1/2)(-3/2)}{2!}(h/x)^2-\ldots)-1}{h}#

#=1/\sqrtx\lim_{h\to 0}\frac{-1/2(h/x)+\frac{(-1/2)(-3/2)}{2!}(h/x)^2-\ldots}{h}#

#=1/\sqrtx\lim_{h\to 0}(-1/2(1/x)+\frac{(-1/2)(-3/2)}{2!}(h/x^2)-\ldots)#

#=1/\sqrtx(-1/2(1/x)+0)#

#=-1/{2x\sqrtx}#

#=-1/{2x^{3/2}}#