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

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

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

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

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

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

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

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