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across Group TitleBest ResponseYou've already chosen the best response.2
\[f(x)=5\sqrt{\ln x}\]\[g(x)=\sqrt{x}\]\[h(x)=\ln x\]\[f(x)=g(h(x))\]\[g'(x)=\frac{1}{2\sqrt{x}}\]\[h'(x)=\frac{1}{x}\]\[f'(x)=g'(h(x))h'(x)=\frac{1}{2x\sqrt{\ln x}}\]
 2 years ago

across Group TitleBest ResponseYou've already chosen the best response.2
I left out the \(5\), but you know what I mean.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
but what if they mean\[\large \sqrt[5]{\ln x}\]?
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
oh no yeah it's 5th root
 2 years ago

across Group TitleBest ResponseYou've already chosen the best response.2
Then all he has to do is change \(g\) and \(g'\) a little.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
oh good, I got scared for a minute :)
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
and don't use substitution for the composite functions just go at it
 2 years ago

Will! Group TitleBest ResponseYou've already chosen the best response.1
\[f(x) = (\lnx)^{1 \over 5}\]\[\ln x = u \rightarrow f(x) = u^{1 \over 5}\]\[{df \over du} = {1 \over 5}u^{4 \over 5} \]\[{du \over dx} = {1 \over x}\] \[{df \over dx} = {df \over du}{du \over dx} = {1 \over x} {{1 \over 5}(\ln x)^{4/5}} = {1 \over 5x(lnx)^{4 \over 5}} \]
 2 years ago
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