What's the derivative of ln(4+x^2*2^-x)
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What's the derivative of ln(4+x^2*2^-x)

[From: ] [author: ] [Date: 12-01-05] [Hit: ]
Based upon that assumption,tAKE THAT, SIMPLIFY IT OUT, AND MULTIPLY IT BACK TO 1/ln(4+x^2*2^-x) and that should give you your answer!......
As a general comment,
d/dx ln[g(x)]=g'(x)/g(x).

Carry this out and you'll have your result, which should be:

-x*(-2+x*ln(2))/[4*2^x+x^2].

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First, we do the regular natural log derivative:

1/ln(4+x^2*2^-x)

But we have to multiply according to the chain rule.

So we pull out (4 + x^2 * 2^-x) and differentiate it.

Idk if you mean [(4+x^2)*(2^-x)] or [4+(x^2 * 2^-x)], but I'll assume it is the former...

Based upon that assumption, we continue to see that the derivative of that is product rule:

(2x)(2^-x) + (4+x^2)[(2^-x) * (-ln2)]

tAKE THAT, SIMPLIFY IT OUT, AND MULTIPLY IT BACK TO 1/ln(4+x^2*2^-x) and that should give you your answer! :)
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