Question

Another I am struggling with...

Differentiate: f(x) = (6-xe^x)/(x+e^x)

Answer 1

pain in the butt quotient rule. be patient and you will get it. use
\[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=6-xe^x,f'(x)=-e^x-xe^x=-e^x(1+x)\]
\[g(x)=x+e^x,g'(x)=1+e^x\]and grind it out

Answer 2

Thanks! I had forgotten that derivative of e^x is e^x

Answer 3

use the quotient rule

(x + e^x)* d(6-xe^x)/dx - (6- xe^x)(1 + e^x)
--------------------------------------
(x + e^x)^ 2

d(6-xe^x)/dx = -xe^x + - e^x

(x + e^x)* (-xe^x + - e^x ) - (6- xe^x)(1 + e^x)
--------------------------------------
(x + e^x)^ 2

= -x^2e^x - xe^x - xe^2x - e^2x - (6 + 6e^x - xe^x - xe^2x)
----------------------------------------------
(x + e^x)^ 2

= -x^2e^x - xe^x - xe^2x - e^2x - 6 - 6e^x + xe^x + xe^2x
----------------------------------------------------
(x + e^x)^ 2


= -x^2e^x - e^2x - 6 - 6e^x
-----------------------
(x + e^x)^ 2


= - e^x(x^2 + e^x + 6) - 6
--------------------
(x + e^x)^ 2

phew that was a grind - satellite was right