# Chain rule of Sin(sqrt(1+x^2)

• Jan 26th 2013, 01:11 PM
mathisfun26
Chain rule of Sin(sqrt(1+x^2)
K so here's the deal my final answer doesn't match up with my book's answer I get ((xcos)(sin)(sqrt(1+x^2)/sqrt(1+x^2)
the answer is the same except sin isn't there. Does any know what I might have done wrong I used the chain rule. Thanks
• Jan 26th 2013, 01:16 PM
Siron
Re: Chain rule of Sin(sqrt(1+x^2)
$\displaystyle \frac{d}{dx}\left[\sin\left(\sqrt{1+x^2}\right)\right] = \cos\left(\sqrt{1+x^2}\right)$$\displaystyle \frac{d}{dx}\left(\sqrt{1+x^2}\right)$$\displaystyle =\frac{\cos\left(\sqrt{1+x^2}\right)2x}{2\sqrt{1+x ^2}}$
$\displaystyle =\frac{x\cos\left(\sqrt{1+x^2}\right)}{\sqrt{1+x^2 }}$
• Jan 26th 2013, 01:21 PM
Plato
Re: Chain rule of Sin(sqrt(1+x^2)
Quote:

Originally Posted by mathisfun26
K so here's the deal my final answer doesn't match up with my book's answer I get ((xcos)(sin)(sqrt(1+x^2)/sqrt(1+x^2)
the answer is the same except sin isn't there. Does any know what I might have done wrong I used the chain rule. Thanks

$\displaystyle \frac{x\cdot\cos(\sqrt{1+x^2})}{\sqrt{1+x^2}}$

• Jan 26th 2013, 01:34 PM
mathisfun26
Re: Chain rule of Sin(sqrt(1+x^2)
My answer is that but with sin I'll post my scratch work in a bit
• Jan 26th 2013, 01:44 PM
Plato
Re: Chain rule of Sin(sqrt(1+x^2)
Quote:

Originally Posted by mathisfun26
My answer is that but with sin I'll post my scratch work in a bit

But you must differentiate $\displaystyle \sin(u)$ to get $\displaystyle \cos(u)du.$
• Jan 26th 2013, 06:54 PM
HallsofIvy
Re: Chain rule of Sin(sqrt(1+x^2)
Quote:

Originally Posted by mathisfun26
My answer is that but with sin I'll post my scratch work in a bit

"That but with sin", or as you first posted it, "(sin)" makes no sense. Sine is a function and must be applied to some variable.