$\displaystyle f(x)=ln(x^3-6)^7$
My professor got
$\displaystyle f'(x)=\frac{21x^2}{x^3-6}$
I have
$\displaystyle f'(x)=\frac{21x^2ln(x^3-6)^6}{x^3-6}$
I don't see how my answer is incorrect
$\displaystyle f(x)=ln(x^3-6)^7$
My professor got
$\displaystyle f'(x)=\frac{21x^2}{x^3-6}$
I have
$\displaystyle f'(x)=\frac{21x^2ln(x^3-6)^6}{x^3-6}$
I don't see how my answer is incorrect
no, it's the argument $\displaystyle (x^3+6)$ that is raised to the 7th power, not the entire log function ... if that were the case, the function would be written $\displaystyle [\ln(x^3+6)]^7$ or the more archaic $\displaystyle \ln^7(x^3+6)$