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Thread: Calc chain rule HELP! Please!

  1. #1
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    Calc chain rule HELP! Please!

    Hey! Doin my math homework and god stuck on one problem ! PLease help:

    Find derivative of:

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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by zaurm View Post
    Hey! Doin my math homework and god stuck on one problem ! PLease help:

    Find derivative of: $\displaystyle (x^2+x+2)^{\sin x}$
    Hint 1:

    Rewrite it as $\displaystyle e^{\ln((x^2+x+2)^{\sin x})} = e^{\sin x(\ln(x^2+x+2))}$.

    Now use the chain rule.

    Hint 2:
    Spoiler:

    The derivative is:

    $\displaystyle e^{\sin x(\ln(x^2+x+2))}*\frac{d}{dx}[\sin x(\ln(x^2+x+2))]$

    $\displaystyle \frac{d}{dx}[\sin x(\ln(x^2+x+2))] = \sin x*\frac{d}{dx}[\ln(x^2+x+2)] + \ln(x^2+x+2)*\frac{d}{dx}[\sin x]$


    Solution:
    Spoiler:

    The derivative is:

    $\displaystyle e^{\sin x(\ln(x^2+x+2))}*\frac{d}{dx}[\sin x(\ln(x^2+x+2))]$

    $\displaystyle \frac{d}{dx}[\sin x(\ln(x^2+x+2))] = \sin x*\frac{d}{dx}[\ln(x^2+x+2)] + \ln(x^2+x+2)*\frac{d}{dx}[\sin x]$

    $\displaystyle = \sin x* \frac{2x+1}{x^2+x+2} + \ln(x^2+x+2)*\cos x = \frac{(2x+1)\sin x}{x^2+x+2} + \ln(x^2+x+2)\cos x$

    So the whole derivative is (bearing in mind that $\displaystyle e^{\sin x(\ln(x^2+x+2))} = (x^2+x+2)^{\sin x}$):

    $\displaystyle (x^2+x+2)^{\sin x}\left(\frac{(2x+1)\sin x}{x^2+x+2} + \ln(x^2+x+2)\cos x\right)$

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