1. ## Arctan Derivitive Question

2. Rearrange

$\displaystyle \tan y=\sqrt{5x^2-1}\implies y'\sec^2y=\frac{5x}{\sqrt{5x^2-1}}$

Make $\displaystyle y'$ the subject of the equation.

3. I need the answer in the form " dy/dx = ", and quite frankly I have no idea how to go about doing this. If someone can post it I'd be grateful.

4. 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?

Differentiate: .$\displaystyle y \;=\;\tan^{-1}\left(5x^2-1\right)^{\frac{1}{2}}$

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: .$\displaystyle \frac{dy}{dx}\;=\;\frac{5x}{5x^2\sqrt{5x^2-1}} \;=\;\frac{1}{x\sqrt{5x^2-1}}$

5. 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: .$\displaystyle \frac{dy}{dx}\;=\;\frac{5x}{5x^2\sqrt{5x^2-1}} \;=\;\frac{1}{x\sqrt{5x^2-1}}$

the poster wants dx/dy

6. Thank you guys so much, the post above was meant to say " dy / dx ." Sorry for the mistype, Jhevon. Thanks again!