Differentiate exponential/logarithm function Given the function, $e^{2x+1}\ln3x$,
I can differentiate the two separately but how do I combine them?
$\dfrac{d}{dx} e^{2x+1}$ = $2e^{2x+1}$
$\dfrac{d}{dx}\ln3x$ = $\ln3$
 A: The Product Rule for Differentiation states that if two functions $f$ and $g$ are differentiable, then their product $fg$ is differentiable and 
$$\bbox[5px,border:2px solid #C0A000]{\frac{d(fg)}{dx}=f\frac{dg}{dx}+g\frac{df}{dx}}$$
Here, we have $f=e^{2x+1}$ and $g=\log 3x$.  Thus, 
$$\frac{df}{dx}=\frac{de^{2x+1}}{dx}=2e^{2x+1}$$
and
$$\frac{dg}{dx}=\frac{d\log 3x}{dx}=\frac{1}{x}$$
The derivative of the product is 
$$\begin{align}
\frac{d(fg)}{dx}&=f\frac{dg}{dx}+g\frac{df}{dx}\\\\
&=\frac{d\left(e^{2x+1}\log 3x\right)}{dx}\\\\
&=\left(e^{2x+1}\right)\times\left(\frac{1}{x}\right)+\left(\log 3x\right)\times \left(2e^{2x+1}\right)
\end{align}$$
A: Using the chain rule, you differentiate each function in turn:
$$\frac{d}{dx}uv = u\frac{d}{dx}v + v\frac{d}{dx}u$$
$$\frac{d}{dx}e^{2x+1}\ln3x = 2e^{2x+1}\ln3x + \frac{1}{x}e^{2x+1}$$
$$\frac{d}{dx}e^{2x+1}\ln3x = e^{2x+1}\left(2\ln3x + \frac{1}{x}\right)$$
Note that the derivate of $\ln3x \neq \ln3 $ becaue $\ln3x = \ln3 + \ln x$ and since $\ln3 $ is just a constant, it becomes zero when its defferentiated leaving you with $\frac{d}{dx}\ln x =\frac{1}{x}$
