Tan square root of (x^2 + 1)
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Originally Posted by henri Tan square root of (x^2 + 1) $\displaystyle \frac{d}{dx}\bigg[\tan\left(\sqrt{x^2+1}\right)\bigg]=\sec^2\left(\sqrt{x^2+1}\right)\cdot\left(\sqrt{x ^2+1}\right)'=\frac{\sec^2\left(\sqrt{x^2+1}\right )x}{\sqrt{x^2+1}}$
note that Mathstud used the chain rule. be sure to look it up and familiarize yourself with it
thank you, I actually had the answer but was not sure
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