## anonymous 5 years ago Find the derivative with respect to x: y=e^(3x) ln (x^2 - 4x) ?

1. anonymous

Use the product rule again:$d(u*v)/dx = u*v' + v*u' = e^{3x}*(2x-4)/(x^2-4x) + \ln(x^2-4x)*(3*e^{3x})$ with the derivative of ln(f(x)) being f'(x)/f(x).

2. anonymous

Ive tried this again too, and can't seem to get the correct answer :(

3. anonymous

:/ Seems my equation was cut off. Lemme re-type it: $dy/dx = e^{3x}*(2x-4)(x^2-4x) + \ln(x^2-4x)*3*e^{3x}$