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Mathematics 20 Online
OpenStudy (anonymous):

derivative y=tan^-1√3x

OpenStudy (anonymous):

\[\frac{d}{dx}\tan^{-1}(\sqrt{3x})\] like that?

OpenStudy (anonymous):

i mean is the radical over all of the inside piece? \(\sqrt{3x}\)??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

if there is a part there that is not clear, let me know

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

yw

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