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I came across this question in my homework and am unsure why it works this way.

Given $y= \ln(e^{x^2})$, find the derivative.

The given answer work showed the formula rewritten as $y=x^{2}$ before starting the differentiation process. My thinking is because

$$f(x)=\ln(x)$$

And the inverse of the natural log function is $$ f^{-1}(x)=e^x$$

Am I right in thinking that multiplying the inverses cancel each other out? If so, why doesn't the x be removed, leaving the 2 as a constant?

J. M. ain't a mathematician
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Jason
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  • I'm not sure I know what you mean by the last sentence. – Qiaochu Yuan Jul 17 '11 at 00:50
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    You seem to be confused with the (admittedly) confusing use of the term "inverse" in mathematics; inverse here means that $\ln(e^x)=x$, not $e^x\ln,x=1$. You can do the simplification suggested, or use the chain rule if you've the discipline. – J. M. ain't a mathematician Jul 17 '11 at 00:52
  • @ J.M: You should put that as an answer. I actually had that written down in my notes, but it was surrounded with other descriptions and I missed it. Thanks! @Quaochu Yuan, JM hit my confusion on the head, check his comment. – Jason Jul 17 '11 at 00:55
  • Why do you keep saying "cancel"? $\ln$ "undoes" what the exponential function does to a number (that precisely is the meaning of "inverse" here). – J. M. ain't a mathematician Jul 17 '11 at 00:59

2 Answers2

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First, you are right that $x\mapsto\ln(x)$ and $x\mapsto e^x$ are inverse functions for each other. This is the most important thing here. For real $x$ we have $\ln(e^x)=x$ and for real $x>0$ we have $e^{\ln(x)}=x$. (since $\ln(x)$ is not defined when $x\leq 0$)

But there is no reason for the $x$ to go away. I'll write it another way: No matter what the box $\square$ is, in the real case we will always have $$\ln (e^{\square})=\square.$$ So for your problem, imagine that $\square=x^2$. Then we must have $$\ln (e^{x^2})=x^2.$$ Lets do more examples:

$$\ln (e^{\sin(x)})=\sin (x)$$

$$\ln(e^{f(x)})=f(x).$$

I hope that helps explain it.

J. M. ain't a mathematician
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Eric Naslund
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    Thanks for the clarification. – Jason Jul 17 '11 at 01:02
  • Eric: I know you know the meaning of what you write but, considering the discussion in the comments about the word inverse, I would suggest replacing the first sentence of your post by something like the functions $x\mapsto\ln(x)$ and $x\mapsto\mathrm{e}^x$ are inverse of each other. – Did Jul 17 '11 at 08:48
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Beware of the unmodified word inverse. The natural log and the exponential functions are compositional inverses of one another. The numbers 2 and 1/2 are multiplicative inverses. Do not confuse these.

ncmathsadist
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  • That clarification helps because I was under the impression that natural log and exponential functions were also multiplicative inverses. – Jason Jul 17 '11 at 12:37