d/dx |x+2|^x (solution known, need explanation)

[Dafydd]

1. The problem statement, all variables and given/known data

I need someone to explain this:
http://www.dafydd.se/stuff/solvethis.png

2. Relevant equations

I guess the following are of relevance...

\frac{d}{dx} |x| = \frac{x}{|x|}
\frac{d}{dx} a^x = a^x ln a

3. The attempt at a solution

The solution is up there. I just need help understanding why this is the solution.

Normally, iirc a derivative of an expression is the outer derivative times the inner derivative, but here it seems to be the original expression times the inner plus the outer derivative, or something in between... eh.

[Cyosis]

Your formula for the derivative of the exponent does not apply here. Note that the a in your formula is a constant, where as the 'a' in this problem is x dependent.

Realise that:

|x+2|^x=e^{\log|x+2|^x}=e^{x \log|x+2|}


Can you take it from here?

[Dafydd]

Hm.


\frac{d}{dx} e^x = e^x \frac{d}{dx} x


and


\frac{d}{dx} ln|x| = \frac{1}{x}


so


\frac{d}{dx} e^{x \log|x+2|} = e^{x \log|x+2|} \frac{d}{dx}(x \log|x+2|) = e^{x \log|x+2|} (ln|x+2| + \frac{x}{x+2}) = |x+2|^x (ln|x+2| + \frac{x}{x+2})


Thanks for the tip, I would never have thought of that. There's too many "tricks" to remember.

[CompuChip]

What you wrote in the last line of your post is correct, I have just a small comment on the notation in the first formula: actually

\frac{d}{dx} e^x = e^x.


What you meant was

\frac{d}{dx} e^{f(x)} = e^{f(x)} \frac{d}{dx} f(x)

which is a consequence of the chain rule. It is this you are using (for f(x) = x log|x + 2|).

You applied it correctly though.

[Dafydd]

Ok, yes, you're right. Thanks!

Not that it matters that much to me right now, but how is it a consequence of the chain rule?

[Cyosis]

If we have a composite function f(g(x)) then the chain rule says that:


\frac{df(g(x))}{dx}=\frac{df}{dg}\frac{dg}{dx}=f'( g(x))g'(x)


The same happens with the exponent where f(x)=e^x, g(x)=x \log|x+2| and f(g(x))=\exp(x \log|x+2|). This is the chain rule.

[g_edgar]

Instead of trying to do derivative formulas with absolute values in them, it may be better to take two cases, x+2 > 0 and x+2 < 0 ... also note that the answer is undefined when x+2 = 0 anyway.

[Dafydd]

Cyosis - got it, thanks!

g_edgar - I agree, but the example was taken from an old exam and what I wrote in the first post was the only answer or explanation given in the solutions. I don't know if a solution taking two cases would have given full credit.