How do I find the derivative

**henri** · October 29th 2008, 05:31 PM

Tan square root of (x^2 + 1)

**Mathstud28** · October 29th 2008, 05:34 PM

Originally Posted by henri

Tan square root of (x^2 + 1)

$\frac{d}{dx}\bigg[\tan\left(\sqrt{x^2+1}\right)\bigg]=\sec^2\left(\sqrt{x^2+1}\right)\cdot\left(\sqrt{x ^2+1}\right)'=\frac{\sec^2\left(\sqrt{x^2+1}\right )x}{\sqrt{x^2+1}}$

**Jhevon** · October 29th 2008, 05:35 PM

note that Mathstud used the chain rule. be sure to look it up and familiarize yourself with it

**henri** · October 29th 2008, 05:54 PM

thank you, I actually had the answer but was not sure

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