# Math Help - Finding a Derivative

1. ## Finding a Derivative

Differentiate.

$f(x) = \frac {7 - xe^x}{x + e^x}$

Sorry I am asking so many questions tonight, but I've been really sick for the past 10 days and have not been able to attend any classes. I'm trying to catch up, but some of these questions seem impossible.

I thought I knew how to do this one. I used the quotient rule, $f'(\frac {u}{v}) = \frac {vu' - uv'}{v^2}$ but I end up getting the wrong answer.

These are my steps:

$\frac {(x+e^x)(7-xe^x)' - (7-xe^x)(x+e^x)'}{(x+e^x)^2}$

I then applied the sum and difference rules to get:

$\frac {(x+e^x)(7'-[xe^x]') - (7-xe^x)(x'+[e^x]')}{(x+e^x)^2}$

2. $f(x)=\frac{7-xe^x}{x+e^x}$

$\frac{dy}{dx}=\frac{(x+e^x)(-xe^x-e^x)-(7-xe^x)(1+e^x)}{(x+e^x)^2}$
If you expand and collect like terms you get...

$\frac{dy}{dx}=\frac{-x^2e^x-e^{2x}-7e^{2x}-7}{(x+e^x)^2}$

3. Originally Posted by topher0805
Differentiate.

$f(x) = \frac {7 - xe^x}{x + e^x}$

Sorry I am asking so many questions tonight, but I've been really sick for the past 10 days and have not been able to attend any classes. I'm trying to catch up, but some of these questions seem impossible.

I thought I knew how to do this one. I used the quotient rule, $f'(\frac {u}{v}) = \frac {vu' - uv'}{v^2}$ but I end up getting the wrong answer.

These are my steps:

$\frac {(x+e^x)(7-xe^x)' - (7-xe^x)(x+e^x)'}{(x+e^x)^2}$
And a good step it is too. To continue:

$= \frac {(x+e^x)(-e^x - xe^x) - (7-xe^x)(1 + e^x)}{(x+e^x)^2}$

where to get $(7 - x e^x)'$ I used the product rule to differentiate $x e^x$ (note that the derivative of 7 is equal to zero). I hope getting $(x+e^x)'$ is no bother for you .....

$= \frac {-(x+e^x)(e^x + xe^x) - (7-xe^x)(1 + e^x)}{(x+e^x)^2}$

Now it's simply a matter of expanding:

$= \frac {-x e^x -x^2 e^x - e^{2x} - x e^{2x} - 7 - 7 e^x + x e^x + x e^{2x}}{(x+e^x)^2}$

and simplifying:

$= \frac {-x^2 e^x - e^{2x} - 7 - 7 e^x}{(x+e^x)^2}$.

From here a variety of re-arrangements are possible .....

(You should check in case I've made any careless mistakes).

4. Thanks for the help. I realized my mistake as I was typing out my steps.

5. Originally Posted by topher0805
Thanks for the help. I realized my mistake as I was typing out my steps.
You're welcome. Hope you're feeling much better. Good luck with it all.