If f(x) = e^[3ln(x^2)], then f'(x) = ?

Question

anonymous · Answer

Using the chain rule a few times, I get the following
f'(x) = e^[3ln(x^2)] * 3x^-2 * 2x = [6e^(3ln(x^2))] / x
But the answer is 6x^5

zepdrix · Answer

I would recommend applying some rules of logarithms before taking the derivative :) It will make this problem a bit easier.

zepdrix · Answer

$$\huge a \log (b)=\log(b^a)$$We can use this to deal with the 3, moving it inside of the log.

zepdrix · Answer

The reason we would want to do that.... is becauseeeee..

The exponential e, and the logarithm are INVERSE operations of one another. So they essentially "cancel out".

Here is a quick example:$$\huge e^{(\ln x)}=x$$

zepdrix · Answer

$$\large e^{3 \ln (x^2)}\qquad \rightarrow \qquad e^{\ln ((x^2)^3)}\qquad \rightarrow \qquad e^{\ln(x^6)}\qquad \rightarrow \qquad x^6$$See what we did there? :o

zepdrix · Answer

That was not the differentiation, that was just simplifying before taking a derivative. Hopefully that makes sense :3
$$\large f(x)=x^6$$

anonymous · Answer

Thank you very much! I completely forgot that e^lnx = x.