Question

How to het f'(x) using the quotient rule on this function: 2x-3 divided by x^2.5?

`(2x-3)/(2x^2.5)`

Answer 1

`f'(x)=-3/2*x^(-5/2)+15/4*x^(-7/2)`

Explanation:

Writing your `f(x)` in the form
`f(x)=(2x)/(2*x^(5/2))-3/(2*x^(5/2))`
and this is

`f(x)=x^(-3/2)-3/2*x^(-5/2)`
so
`f'(x)=-3/2x^(-5/2)+15/4*x^(-7/2)`

Answer 2

`f'(x) = frac{15sqrtx - 6x^(3/2)}{4x^4}`

Explanation:

`f(x) = frac{2x-3}{2x^(5/2)}`

Note that the quotient rule states that if `color(blue)(f(x) = g(x)/(h(x)))`, then `color(blue)(f'(x) = frac{h(x)g'(x)-g(x)h'(x)}{(h(x))^2})`.

We apply the quotient rule to this context:

`f'(x) = frac{(2x^(5/2))(2)-(2x-3)(5/2 * 2 * x^(3/2))}{(2x^(5/2))^2}`

Simplify:
`f'(x) = frac{4x^(5/2) - (2x-3)(5x^(3/2))}{4x^5}`

Simplify further by distributing:
`f'(x) = frac{4x^(5/2) - 10x^(5/2) + 15x^(3/2)}{4x^5}`

`f'(x) = frac{-6x^(5/2) + 15x^(3/2)}{4x^5}`

`f'(x) = frac{x(-6x^(3/2) + 15x^(1/2))}{4x^5}`

`f'(x) = frac{15sqrtx - 6x^(3/2)}{4x^4}`