Derivative of natural log

**amthomasjr** · October 31st 2012, 01:18 PM

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

My professor got

$f'(x)=\frac{21x^2}{x^3-6}$

I have

$f'(x)=\frac{21x^2ln(x^3-6)^6}{x^3-6}$

I don't see how my answer is incorrect

**skeeter** · October 31st 2012, 01:33 PM

note the power rule for logs ...

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

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

$f'(x) = 7 \cdot \frac{3x^2}{x^3-6}$

**amthomasjr** · October 31st 2012, 01:41 PM

I thought that, but wouldn't the parentheses make a difference?

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

but $ln(a)^b\neq bln(a)$ ?

**skeeter** · October 31st 2012, 02:22 PM

no, it's the argument $(x^3+6)$ that is raised to the 7th power, not the entire log function ... if that were the case, the function would be written $[\ln(x^3+6)]^7$ or the more archaic $\ln^7(x^3+6)$

**amthomasjr** · October 31st 2012, 02:53 PM

I see. I've always thought that $[ln(x^3-6)]^7$ was implied when you write it as $ln(x^3-6)^7$ . Thanks.

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