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The original question is $f(x) = 7\ln(5xe^{-x})$

I'm not sure if I have to use the chain rule to figure out $\ln(5xe^{-x})$ because $5xe^{-x}$ is one term within ln.

My guess is that it's like this:

$$7(-(e^{-x-1})/e^{-x})$$

or just simply $-7$.

I'm specifically unsure with how to find the derivative of $5xe^{-x}$.

I know that $e^x$'s derivative is simply $e^x(1)$ because the derivative of $x = 1$

so when I find the derivative of $e^{-x}$ I'd expect it to be $-1e^x$ and in my case $-5xe^{-x}$

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Simplify the logarithm BEFORE differentiating. You'll get $7\ln5 + 7\ln x - 7x$. Then differentiate. –  Michael Hardy Jun 5 '12 at 22:59
    
I see. This is what DonAntonio said. You crystalized it for me. Thanks! –  ninja08 Jun 5 '12 at 23:01

2 Answers 2

up vote 6 down vote accepted

Hint:$$\ln\left(5xe^{-x}\right)=\ln 5+\ln x+\ln e^{-x}\ldots$$

Now just remember that the natural log. and the exponential function are inverse to each other and also $\,\left(\ln x\right)'=\frac{1}{x}\,$

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I didn't catch that I could simplify it. Thanks! –  ninja08 Jun 5 '12 at 23:01

Let $g(x)=5xe^{-x}$, then $f'(x)=7\frac{g'(x)}{g(x)}$ by the derivative rule for $\ln$. Then use the product rule to find $g'(x)$. In fact, $$ g'(x)=5(e^{-x}+x(-e^{-x}))=5(e^{-x}-xe^{-x}). $$ Resubstitute everything to get your answer in terms of $x$.

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