Question

Help with this calculus problem.

Find the derivative of y=(tan^-1)(1/2x)

thanks!

Answer 1

is it
\[\tan^{-1}(\frac{x}{2})\]

Answer 2

\[\tan^{-1} (1/2x)\]

Answer 3

use chain rule.
\[\tan^{-1}(\frac{1}{2x})\]?

Answer 4

\[\frac{d}{dx}\tan^{-1}(x)=\frac{1}{x^2+1}\] and by the chain rule
\[\frac{d}{dx}\tan^{-1}(\frac{1}{2x})=\frac{1}{(\frac{1}{2x})^2+1}\times \frac{d}{dx}\frac{1}{2x}\]

Answer 5

\[\frac{d}{dx}\frac{1}{2x}=-\frac{1}{2x^2}\] so your initial answer is
\[\frac{d}{dx}\tan^{-1}(\frac{1}{2x})=\frac{1}{(\frac{1}{2x})^2+1}\times \frac{-1}{2x^2}\] then a ton of algebra to clean this up

Answer 6

good from here?

Answer 7

Yeah thanks!