Question

How do you use the definition of a derivative to find the derivative of `(2/sqrt x)`?

Answer 1

`f'(x)=1/(xsqrtx)=1/x^(3/2)`

Explanation:

The limit definition of a derivative states that for a function `f(x)`, its derivative is

`f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h`

Here, `f(x)=2/sqrtx`, so

`f'(x)=lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h`

Multiply the entire fraction by `sqrt(x+h)sqrtx`.

`f'(x)=lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h((sqrt(x+h)sqrtx)/(sqrt(x+h)sqrtx))`

`f'(x)=lim_(hrarr0)(2sqrtx-2sqrt(x+h))/(hsqrt(x+h)sqrtx)`

`f'(x)=lim_(hrarr0)(2(sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrtx)`

Multiply by the conjugate of the term in the numerator.

`f'(x)=lim_(hrarr0)(2(sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrtx)((sqrtx+sqrt(x+h))/(sqrtx+sqrt(x+h)))`

`f'(x)=lim_(hrarr0)(2(x-(x-h)))/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))`

`f'(x)=lim_(hrarr0)(2h)/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))`

`f'(x)=lim_(hrarr0)2/(sqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))`

Evaluate the limit by plugging in `0` for `h`.

`f'(x)=2/(sqrt(x+0)sqrtx(sqrtx+sqrt(x+0)))`

`f'(x)=2/(sqrtxsqrtx(sqrtx+sqrtx)`

`f'(x)=2/(x(2sqrtx))`

`f'(x)=1/(xsqrtx)=1/x^(3/2)`