Question

Derivative (Quotient Rule)
f(x)= x^2/((squared root x)-3)

Answer 1

I got it to here:
\[\frac{ \frac{ 3 }{ 2 }x ^{3/2}-6x }{ (x ^{1/2}-3)^2 }\]
I'm not sure if it's right

Answer 2

[2x* (sqrt(x)-3)-x^2*(1/2(x)^(-1/2)]/(sqrt(x)-3)^2

Answer 3

\[f(x) = \frac{x^2}{x^\frac{1}{2}-3}\]\[\text{Let } g(x) = x^2, h(x) = x^\frac{1}{2}-3\]
What are g'(x) and h'(x)?

Answer 4

g'(x) = 2x
h'(x)=1/2x^-1/2

Answer 5

Yup!
\[f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{[h(x)]^2} = \frac{(x^\frac{1}{2}-3)(2x)-x^2(\frac{1}{2}x^{-\frac{1}{2}})}{(x^\frac{1}{2}-3)^2}\]

Answer 6

\[= \frac{2x^\frac{3}{2}-6x-\frac{1}{2}x^{\frac{3}{2}}}{(x^\frac{1}{2}-3)^2} = \frac{\frac{3}{2}x^\frac{3}{2}-6x}{(x^\frac{1}{2}-3)^2} \]
Your answer looks good to me.

Answer 7

but at the back of the textbook it says:
\[f'(x)=\frac{ 3x \sqrt{x}-12x }{ 2(\sqrt{x}-3)^2 }\]

Answer 8

LOL, they are just the same!

\[\frac{\frac{3}{2}x^\frac{3}{2}-6x}{(x^\frac{1}{2}-3)^2} = \frac{\frac{1}{2}(3x^\frac{3}{2}-12x)}{(x^\frac{1}{2}-3)^2} = \frac{3x^\frac{3}{2}-12x}{2(x^\frac{1}{2}-3)^2} =\frac{3x(x^\frac{1}{2})-12x}{2(x^\frac{1}{2}-3)^2} = \frac{3x\sqrt x-12x}{2(x^\frac{1}{2}-3)^2}\]

Answer 9

Wow! Did not know that. Thanks for writing that for me! :)

Answer 10

Welcome :)