Differentiate the function? f(x) = log13(xe^x)

Mar 17, 2017

$f ' \left(x\right) = {e}^{x} \log 13 \left(1 + x\right)$

Explanation:

$f \left(x\right) = \log 13 \left(x {e}^{x}\right)$

Recall that the product rule allows us to differentiate a function that is the product of two functions.

If $h \left(x\right) = f \left(x\right) \cdot g \left(x\right)$
then $h ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

So, if we have

$f \left(x\right) = x {e}^{x} \log 13$

Using the product rule:

$f ' \left(x\right) = \left(1\right) \log 13 {e}^{x} + x {e}^{x} \log 13$

$f ' \left(x\right) = {e}^{x} \log 13 \left(1 + x\right)$

Mar 17, 2017

As presented, we have $f \left(x\right) = \log \left(13 x {e}^{x}\right)$. I'll assume that $\log$ is natural logarithm.

Explanation:

$\frac{d}{\mathrm{dx}} \left(\log u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$

So we have $f ' \left(x\right) = \frac{1}{13 x {e}^{x}} \frac{d}{\mathrm{dx}} \left(13 x {e}^{x}\right)$.

Using the product rule we get

$\frac{d}{\mathrm{dx}} \left(13 x {e}^{x}\right) = 13 {e}^{x} + 13 x {e}^{x} = 13 {e}^{x} \left(1 + x\right)$.

Thus $f ' \left(x\right) = \frac{1}{13 x {e}^{x}} \cdot 13 {e}^{x} \left(1 + x\right) = \frac{1 + x}{x} = \frac{1}{x} + 1$.

Mar 17, 2017

Perhaps the intended function is $f \left(x\right) = {\log}_{13} \left(x {e}^{x}\right)$ in which case

Explanation:

Use $\frac{d}{\mathrm{dx}} \left({\log}_{b} u\right) = \frac{1}{u \ln b} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ to get

$f ' \left(x\right) = \frac{1}{x {e}^{x} \ln 13} \cdot \frac{d}{\mathrm{dx}} \left(x {e}^{x}\right)$

Now use the product rule to find

$\frac{d}{\mathrm{dx}} \left(x {e}^{x}\right) = {e}^{x} + x {e}^{x} = {e}^{x} \left(1 + x\right)$.

So we have

$f ' \left(x\right) = \frac{1}{x {e}^{x} \ln 13} \cdot {e}^{x} \left(1 + x\right) = \frac{1 + x}{x \ln 13}$