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Math Help - Chain rule of Sin(sqrt(1+x^2)

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

    \frac{d}{dx}\left[\sin\left(\sqrt{1+x^2}\right)\right] = \cos\left(\sqrt{1+x^2}\right) \frac{d}{dx}\left(\sqrt{1+x^2}\right) =\frac{\cos\left(\sqrt{1+x^2}\right)2x}{2\sqrt{1+x  ^2}}
    =\frac{x\cos\left(\sqrt{1+x^2}\right)}{\sqrt{1+x^2  }}
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    Re: Chain rule of Sin(sqrt(1+x^2)

    Quote Originally Posted by mathisfun26 View Post
    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

    I find it hard to read your answer. The correct answer is:
    \frac{x\cdot\cos(\sqrt{1+x^2})}{\sqrt{1+x^2}}

    Is that your answer?
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    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
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    Re: Chain rule of Sin(sqrt(1+x^2)

    Quote Originally Posted by mathisfun26 View Post
    My answer is that but with sin I'll post my scratch work in a bit
    But you must differentiate \sin(u) to get \cos(u)du.
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    Re: Chain rule of Sin(sqrt(1+x^2)

    Quote Originally Posted by mathisfun26 View Post
    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.
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