$\displaystyle f(x)=ln(x^3-6)^7$

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

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- Oct 31st 2012, 01:18 PMamthomasjrDerivative of natural log
$\displaystyle f(x)=ln(x^3-6)^7$

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 - Oct 31st 2012, 01:33 PMskeeterRe: Derivative of natural log
note the power rule for logs ...

$\displaystyle \ln{a^b} = b\ln{a}$

$\displaystyle f(x) = \ln(x^3-6)^7 = 7\ln(x^3-6)$

$\displaystyle f'(x) = 7 \cdot \frac{3x^2}{x^3-6}$ - Oct 31st 2012, 01:41 PMamthomasjrRe: Derivative of natural log
I thought that, but wouldn't the parentheses make a difference?

$\displaystyle ln(a^b)=bln(a)$

but $\displaystyle ln(a)^b\neq bln(a)$? - Oct 31st 2012, 02:22 PMskeeterRe: Derivative of natural log
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 $\displaystyle \ln^7(x^3+6)$

- Oct 31st 2012, 02:53 PMamthomasjrRe: Derivative of natural log
I see. I've always thought that $\displaystyle [ln(x^3-6)]^7$ was implied when you write it as $\displaystyle ln(x^3-6)^7$. Thanks.