Question

please help. derivative of [ln(3x^5)]/(e^2x)

Answer 1

quotient rule. also chain rule. but before you start, make life easier and rewrite
\[\ln(3x^5)\] as \[\ln(3)+5\ln(x)\] so save a lot of trouble and use of the chain rule

Answer 2

then you use
\[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\]with \[f(x)=\ln(3)+5\ln(x), f'(x)=\frac{5}{x}\]
\[g(x) = e^{2x}, g'(x)=2e^{2x}\] and put it in the formula

Answer 3

I got \[[15/3x ^{4}-2\ln(3x ^{5})]/(e ^{2x})\] right?