Question

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

Answer 1

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}}`