Originally Posted by
Soroban Hello, Quiz0n!
You are expected to know that: .$\displaystyle \frac{d}{dx}\!\left(\tan^{\text{-}1}\!u\right) \;=\;\frac{1}{1+u^2}\!\cdot\!\frac{du}{dx}$ . . . and the Chain Rule.
So exactly where is your difficulty?
We have: .$\displaystyle \frac{dy}{dx}\;=\;\frac{1}{1 + \left[\left(5x^2-1\right)^{\frac{1}{2}}\right]^2} \cdot \frac{1}{2}(5x^2-1)^{\text{-}\frac{1}{2}} \cdot 10x \;=\;\frac{1}{1 + 5x^2 - 1}\cdot\frac{5x}{\sqrt{5x^2-1}} $
Therefore: .$\displaystyle \frac{dy}{dx}\;=\;\frac{5x}{5x^2\sqrt{5x^2-1}} \;=\;\frac{1}{x\sqrt{5x^2-1}} $