Use logarithmic differentiation to find the derivative of f. f(x)=(3+x)^(2x) i have 2x ln (3+x) but i don't know where to go from there...
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$\displaystyle y=(3+x)^{2x}$ $\displaystyle ln(y)=2xln(3+x)$ $\displaystyle \frac{y'}{y}=\frac{2x}{x+3}+2ln(x+3)$ Now, solve for y and remember what y equals.
I got y'=(3+x)^(2x)*((2x)/(x+3)+2ln(x+3))
That's it.
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