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].
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! :)
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! :)