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OpenStudy (anonymous):
Find the derivative of the function.
h(x) = log6(x2 + x)
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terenzreignz (terenzreignz):
Seems like a true-blue application of the chain rule?
OpenStudy (anonymous):
that, and the fact that
\[\log_6(x)=\frac{\ln(x)}{\ln(6)}\] so the derivative of
\[\log_6(x)\] is
\[\frac{1}{\ln(6)x}\]
terenzreignz (terenzreignz):
anytime you're ready @arshia93 ^_^
OpenStudy (anonymous):
i put \[\frac{ 1 }{ (2x+1)\ln6 }\] but that came out incorrect.
OpenStudy (anonymous):
@terenzreignz
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terenzreignz (terenzreignz):
Chain rule though.
Bit by bit...
terenzreignz (terenzreignz):
Remember...\[\Large \frac{d}{dx}f(g(x))= f'(g(x)) g'(x)\]
terenzreignz (terenzreignz):
So you could take
\[\Large f(x) = \log_6(x)\]\[\Large g(x) = x^2 +x\]
OpenStudy (anonymous):
im supposed to fin detivative of x^2+x correct?
So i included (2x+1) with the fraction from log6
OpenStudy (anonymous):
is it not included in the fraction?
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terenzreignz (terenzreignz):
Let me be more blunt about it ^_^
\[\Large \frac{d}{dx} \log_6[g(x)]= \frac{g'(x)}{g(x)\ln (6)}\]
PLEASE review the chain rule.
OpenStudy (anonymous):
\[\frac{ 2x+1 }{ \ln6(x^2+x) }?\]
OpenStudy (anonymous):
when you use the chain rule never change the input!!
OpenStudy (anonymous):
yes
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