# Math Help - derivative.. problem

1. ## derivative.. problem

Hello,
I have a quotient problem on the photo attached,
I have gone wrong on some parts involving exponents I think

I need some kind person to show me how to do it PLEASE

2. I strongly suggest that you rewrite the problem at the beginning to make it easier.

$f(x) ~=~ \frac{x^{-1}-2x^{-2}}{2x^{-3}-3x^{-4}}
~=~ \frac{x^{-1}-2x^{-2}}{2x^{-3}-3x^{-4}} \cdot \frac{x^4}{x^4}
~=~ \frac{x^3-2x^2}{2x-3}$

Now, work from this instead and it should be a lot easier.

3. Originally Posted by wolfhound
Hello,
I have a quotient problem on the photo attached,
I have gone wrong on some parts involving exponents I think

I need some kind person to show me how to do it PLEASE

Are you asking to find the derivative of

$f(x) = \frac{\frac{1}{x} - \frac{2}{x^2}}{\frac{2}{x^3} - \frac{3}{x^4}}$?

$f(x) = \frac{\frac{x - 2}{x^2}}{\frac{2x - 3}{x^4}}$

$= \frac{x^2(x - 2)}{2x - 3}$

$= \frac{x^3 - 2x^2}{2x - 3}$.

So now by using the quotient rule...

$f'(x) = \frac{(2x - 3)(3x^2 - 4x) - 2(x^3 - 2x^2)}{(2x - 3)^2}$

$= \frac{6x^3 - 8x^2 - 9x^2 + 12x - 2x^3 + 4x^2}{(2x - 3)^2}$

$= \frac{4x^3 + 13x^2 + 12x}{(2x - 3)^2}$

$= \frac{x(4x^2 + 13x + 12)}{(2x - 3)^2}$

4. Originally Posted by wolfhound
Hello,
I have a quotient problem on the photo attached,
I have gone wrong on some parts involving exponents I think

I need some kind person to show me how to do it PLEASE

Dude, you love to suffer. First, simplify the expression for f(x)(remember: $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$):

$f(x)=\frac{\displaystyle{\frac{1}{x}-\frac{2}{x^2}}}{\displaystyle{\frac{2}{x^3}-\frac{3}{x^4}}}=\frac{\displaystyle{\frac{x-2}{x^2}}}{\displaystyle{\frac{2x-3}{x^4}}}$ $=\frac{x^2(x-2)}{2x-3}=\frac{x^3-2x^2}{2x-3}$ ,and now you derivate: much simpler.

Tonio

5. Thanks for the help everyone
excellent!