Question

How do you use the definition of a derivative to find the derivative of `1/sqrt(x)`?

Answer 1

First, remember that square roots can be rewritten in exponential forms:

`root(n)(x^m)` = `x^(m/n)`

As you have a simple square root in the denominator of your function, we can rewrite it as `x^(1/2)`, alright?

Now, remembering that when potences are on the denominator you can bring them to the numerator by changing its positivity/negativity, you can rewrite `1/(x^(1/2))` as `x^(-1/2)`

Deriving now your function, you'll get:

`-(1/2).x^(-3/2)`

Alternatively: `-1/(2(x^(3/2))`

Or even in this form: `-1/(root(2)(x^3))`

Answer 2

`f(x)=1/sqrtx`

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

`= lim_(hrarr0) (1/sqrt(x+h)-1/sqrt(x))/h`

`= lim_(hrarr0) ((sqrtx-sqrt(x+h))/(sqrt(x+h)sqrt(x)))/(h/1)`

`= lim_(hrarr0) (sqrtx-sqrt(x+h))/(sqrt(x+h)sqrt(x)) * 1/h`

`= lim_(hrarr0) ((sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrt(x)) * ((sqrtx+sqrt(x+h)))/((sqrtx+sqrt(x+h)))`

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

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

`= lim_(hrarr0) (-1)/(sqrt(x+h)sqrt(x)(sqrtx+sqrt(x+h)))`

`= (-1)/(sqrt(x+0)sqrt(x)(sqrtx+sqrt(x+0)))`

`= (-1)/(x(2sqrtx)) =(-1)/(2xsqrtx) = (-1)/(2x^(3/2))`