1. ## Find the Derivative

y=(9+4x-8 sqrt x)/x

Using the quotient rule I get

[x(4-4x^-1/2)-(9+4x-8x^1/2)]x^2

[4x-(4x^1/2)-9-4x+(8x^1/2)]x^2

[(8x^1/2)-9]x^2

However the boosk final answer is

-9/x^2 + 4/xsqrtx

It has been a year since I took trig and I know my algebra is horrible but how did they get 4/xsqrtx from 4sqrtx/x^2

Thanks for the help.

2. ## Re: Find the Derivative

Originally Posted by oriont
y=(9+4x-8 sqrt x)/x

Using the quotient rule I get

[x(4-4x^-1/2)-(9+4x-8x^1/2)]x^2

[4x-(4x^1/2)-9-4x+(8x^1/2)]x^2

[(8x^1/2)-9]x^2

However the boosk final answer is

-9/x^2 + 4/xsqrtx

It has been a year since I took trig and I know my algebra is horrible but how did they get 4/xsqrtx from 4sqrtx/x^2

Thanks for the help.
$\frac{4\sqrt{x}}{x^2}=(4x^{1/2})(x^{-2})=4x^{1/2-2}=4x^{-3/2}=\frac{4}{x\sqrt{x}}$

You have the same answer as the book.

3. ## Re: Find the Derivative

As an alternative, you can use logarithmic differentiation.

$y=\frac{9+4x-8\sqrt{x}}{x}$

$\frac{1}{y}\frac{dy}{dx}=\frac{4+\frac{8}{2\sqrt{x }}}{xy}-\frac{1}{x}$

$\therefore \frac{dy}{dx}=\cdots = \frac{4\sqrt{x}-9}{x^2}$

4. ## Re: Find the Derivative

$y = \frac{9+4x-8\sqrt{x}}{x} = \frac{9}{x} + \frac{4x}{x} - \frac{8\sqrt{x}}{x} = 9x^{-1} + 4 - 8x^{-1/2}$

$y' = -9x^{-2} + 4x^{-3/2} = -\frac{9}{x^2} + \frac{4}{x^{3/2}} = -\frac{9}{x^2} + \frac{4}{x\sqrt{x}}$