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derivative y=tan^-1√3x
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\[\frac{d}{dx}\tan^{-1}(\sqrt{3x})\] like that?
i mean is the radical over all of the inside piece? \(\sqrt{3x}\)??
yes
ok then we need two facts: one fact is \(\frac{d}{dx}\tan^{-1}(x)=\frac{1}{x^2+1}\) the other is \(\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\) and also chain rule
this gives \[\frac{d}{dx}\tan^{-1}(\sqrt{3x})=\frac{1}{(\sqrt{3x})^2+1}\times \frac{\sqrt{3}}{2\sqrt{x}}\] which you can clean up via some algebra
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if there is a part there that is not clear, let me know
thank you!
yw
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