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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?

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    Why do you keep saying "cancel"? $\ln$ "undoes" what the exponential function does to a number (that precisely is the meaning of "inverse" here).2011-07-17

2 Answers 2

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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.

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    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*.2011-07-17
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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.

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    That clarification helps because I was under the impression that natural log and e$x$ponential functions were also multiplicative inverses.2011-07-17