1. Bit of help please, rule

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

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

has derive.

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

Thank you

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

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

has derive.

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

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

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

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

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

has derive.

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

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

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

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

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