Question #a787b

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Question #a787b

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Answer 1 · 2016-10-22T17:53:42

$\frac{d}{d x} (\frac{log (x e^{x})}{log (5)}) = \frac{x + 1}{x log (5)}$ $d/dx(log(xe^x)/log(5))=(x+1)/(xlog(5))$

Explanation:

Another way to do this is to simplify at first using the rule $log (A B) = log (A) + log (B)$ $log(AB)=log(A)+log(B)$ :

$\frac{log (x e^{x})}{log (5)} = \frac{log (x) + log (e^{x})}{log (5)} = \frac{log (x) + x}{log (5)}$ $log(xe^x)/log(5)=(log(x)+log(e^x))/log(5)=(log(x)+x)/log(5)$

Don't forget that $log (5)$ $log(5)$ is a constant, so we can bring it out when differentiating:

$\frac{d}{d x} (\frac{log (x) + x}{log (5)}) = \frac{1}{log (5)} \frac{d}{d x} (log (x) + x)$ $d/dx((log(x)+x)/log(5))=1/log(5)d/dx(log(x)+x)$

The derivative of $log (x)$ $log(x)$ is $\frac{d}{d x} log (x) = \frac{1}{x}$ $d/dxlog(x)=1/x$ . The derivative of $x$ $x$ is $1$ $1$ .

$\frac{1}{log (5)} \frac{d}{d x} (log (x) + x) = \frac{1}{log (5)} (\frac{1}{x} + 1) = \frac{1}{log (5)} (\frac{1 + x}{x})$ $1/log(5)d/dx(log(x)+x)=1/log(5)(1/x+1)=1/log(5)((1+x)/x)$

Combining this all:

$\frac{d}{d x} (\frac{log (x e^{x})}{log (5)}) = \frac{x + 1}{x log (5)}$ $d/dx(log(xe^x)/log(5))=(x+1)/(xlog(5))$

Answer 2 · 2016-10-22T18:15:53

$\frac{1 + x}{x ln 5}$ $(1+x) / (xln5)$

Explanation:

STEP 1: Rewrite the expression using logarithm rules
If we remember our logarithm rules, we will recall that ${log}_{b} x = \frac{log x}{log b}$ $log_bx = logx/logb$ .
Using the logarithm rule above, we can rewrite $\frac{log (x e^{x})}{log 5}$ $log(xe^x)/log5$ as ${log}_{5} (x e^{x})$ $log_5(xe^x)$ .

STEP 2: Differentiate
Recall our differentiation rule for logarithms: $\frac{d}{d x} {log}_{a} x = \frac{1}{x ln a}$ $d/dx log_ax=1/(xlna)$

Combining our log derivative rule with the chain rule, we get:
$\frac{d}{d x} {log}_{5} (x e^{x}) = \frac{1}{(x e^{x}) ln 5} \cdot \frac{d}{d x} (x e^{x})$ $d/dx log_5(xe^x) = 1/((xe^x)ln5) * d/dx(xe^x)$
To take the derivative of $x e^{x}$ $xe^x$ we must remember to apply the product rule:
$\frac{1}{(x e^{x}) ln 5} \cdot \frac{d}{d x} (x e^{x}) = \frac{1}{(x e^{x}) ln 5} \cdot [1 \cdot e^{x} + x \cdot e^{x}]$ $1/((xe^x)ln5) * d/dx(xe^x) = 1/((xe^x)ln5) * [1*e^x + x*e^x]$

STEP 3: Simplify
$\frac{1}{(x e^{x}) ln 5} \cdot [1 \cdot e^{x} + x \cdot e^{x}] = \frac{e^{x} + x e^{x}}{(x e^{x}) ln 5} = \frac{1 + x}{x ln 5}$ $1/((xe^x)ln5) * [1*e^x + x*e^x]=[e^x + xe^x] / ((xe^x)ln5) = (1+x) / (xln5)$

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