Arctan Derivitive Question

**Quiz0n** · October 8th 2007, 11:03 AM

Can someone please help me in finding the derivitive of the function:

Technically it's finding (dy/dx), but please help!

**Krizalid** · October 8th 2007, 11:13 AM

Rearrange

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

Make $y'$ the subject of the equation.

**Quiz0n** · October 8th 2007, 11:32 AM

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.

**Soroban** · October 8th 2007, 11:44 AM

Hello, Quiz0n!

You are expected to know that: . $\frac{d}{dx}\!\left(\tan^{\text{-}1}\!u\right) \;=\;\frac{1}{1+u^2}\!\cdot\!\frac{du}{dx}$ . . . and the Chain Rule.

So exactly where is your difficulty?

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

We have: . $\frac{dy}{dx}\;=\;\frac{1}{1 + \left[\left(5x^2-1\right)^{\frac{1}{2}}\right]^2} \cdot \frac{1}{2}(5x^2-1)^{\text{-}\frac{1}{2}} \cdot 10x \;=\;\frac{1}{1 + 5x^2 - 1}\cdot\frac{5x}{\sqrt{5x^2-1}}$

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

**Jhevon** · October 8th 2007, 11:46 AM

Originally Posted by Soroban

Hello, Quiz0n!

You are expected to know that: . $\frac{d}{dx}\!\left(\tan^{\text{-}1}\!u\right) \;=\;\frac{1}{1+u^2}\!\cdot\!\frac{du}{dx}$ . . . and the Chain Rule.

So exactly where is your difficulty?

We have: . $\frac{dy}{dx}\;=\;\frac{1}{1 + \left[\left(5x^2-1\right)^{\frac{1}{2}}\right]^2} \cdot \frac{1}{2}(5x^2-1)^{\text{-}\frac{1}{2}} \cdot 10x \;=\;\frac{1}{1 + 5x^2 - 1}\cdot\frac{5x}{\sqrt{5x^2-1}}$

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

the poster wants dx/dy

**Quiz0n** · October 8th 2007, 11:53 AM

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

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