Question

differentiate
$y = \ln \frac{(x + 3)^2}{e^{x - 4}}$
$\frac{d}{dx}\left\ln \frac{(x + 3)^2}{e^{x - 4}}\
ight=\square$

Explanation:

Step1: Use logarithm properties

Use $\ln\frac{a}{b}=\ln a-\ln b$ and $\ln a^n = n\ln a$.
$$y=\ln(x + 3)^2-\ln e^{x - 4}=2\ln(x + 3)-(x - 4)$$

Step2: Differentiate term - by - term

Differentiate $2\ln(x + 3)$ using $\frac{d}{dx}\ln u=\frac{u'}{u}$ ($u=x + 3$, $u' = 1$), and differentiate $-(x - 4)$ using $\frac{d}{dx}(ax + b)=a$.
For $y_1 = 2\ln(x + 3)$, $y_1'=\frac{2}{x + 3}$. For $y_2=-(x - 4)=-x + 4$, $y_2'=-1$.

Answer:

$\frac{2}{x + 3}-1$