1. Bit of help please, rule

Ok, so how to do we use the quotient rule to show that this function

$\displaystyle g(x) = \frac{e^{(20-x/36)}}{(5+x)^2}$ ................ $\displaystyle (x>-5)$

has derive.

$\displaystyle k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$

Thank you

Ok, so how to do we use the quotient rule to show that this function

$\displaystyle g(x) = \frac{e^{(20-x/36)}}{(5+x)^2}$ ................ $\displaystyle (x>-5)$

has derive.

$\displaystyle k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$

Thank you
Let $\displaystyle f= e^{20-x/36}$ and let $\displaystyle g=(5+x)^2$ then use the quotient formula $\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{d}{dx}[f(x)]-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2}$ and simplify.
Keep in mind that the derivitive of $\displaystyle (5+x)^2$ is 2(x+5) and $\displaystyle f'(x)= \frac{1}{36}e^{20-x/36}$ (you can verify). So you have:

$\displaystyle \frac{[(5+x)^2 . \frac{e^{20-x/36}}{36}] - [e^{20-x/36} . 2(x+5)]}{(5+x)^4} = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$

Ok, so how to do we use the quotient rule to show that this function

$\displaystyle g(x) = \frac{e^{(20-x/36)}}{(5+x)^2}$ ................ $\displaystyle (x>-5)$

has derive.

$\displaystyle k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$

Thank you
$\displaystyle u = e^{20-x/36}$ $\displaystyle u' = -\frac{1}{36}e^{20-x/36}$

$\displaystyle v = (5+x)^2$ $\displaystyle v' = 2(5+x)$

$\displaystyle g'(x) = \frac{-(5+x)^2 \cdot \frac{1}{36}e^{20-x/36} - 2(5+x)e^{20-x/36}}{(5+x)^4} = -\frac{e^{20-x/36}\left({x} +77\right)}{36(5+x)^3}$

4. looking at both your posts, they are both different solutions arent they?