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Find the derivative of y with respect to x. y=tan^-1 (ln4x)
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\[y= \tan ^{-1}(\ln4x)\]
\[4/4x(1+(\ln4x)^{2})\]
If you have a chance, may you explain a little
The derivative of InvTan is 1/(1 + x^2) du/dx, so 1/(1+ (ln4x)^2), then you take the derivative of ln4x and multiply it by that. The derivative of ln4x is 4/4x, so multiply 4/4x by 1/(1+ (ln4x)^2) and you have your answer
Okay, it's starting to make sense. It's pretty simple then, but I missed it on my test. Thanks for your help
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