What's the derivative of # arctan(x)^(1/2)#?

1 Answer
Eddie
Jun 21, 2016

so #\ y' = \frac{1}{2 \sqrt x} .\frac{1}{x+1 }#

Explanation:

let #\tan y = x^(1/2)#

so #\sec^2 y \ y' = \frac{1}{2 \sqrt x}#

using the #tan^2 + 1 = sec^2# identity, #sec^2 y = tan^2 y + 1 = x + 1#

so #\ y' = \frac{1}{2 \sqrt x} \frac{1}{\sec^2 y } = \frac{1}{2 \sqrt x} .\frac{1}{x+1 }#

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