The following Problem:

$\displaystyle f(x)=5^{x^2+2x}$

Holds a derivative of:

$\displaystyle f'(x)=(2*log(5)*x+2*log(5))*5^{x^2+2*x}$

I need to know how this log came about.

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- Mar 13th 2010, 10:29 PM #1

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- Mar 13th 2010, 10:40 PM #2

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$\displaystyle f(x)=5^{x^2+2x}$

$\displaystyle \ln f(x)=\ln \left(5^{x^2+2x}\right)$

$\displaystyle \ln f(x)=(x^2+2x) \ln 5$

$\displaystyle \frac{d}{dx}\left[\ln f(x)\right]=\frac{d}{dx}\left[(x^2+2x) \ln 5\right]$

$\displaystyle \frac{f'(x)}{f(x)}=(2x+2) \ln 5$

$\displaystyle f'(x)=f(x) (2x+2) \ln 5$

$\displaystyle f'(x)= 5^{x^2+2x} (2x+2) \ln 5$

Of course, when calculating a derivative like this, you wouldn't actually go through all these steps. Instead, you can just use the well known rule:

$\displaystyle \frac{d}{dx} a^{f(x)} = a^{f(x)} f'(x) \ln a$