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Thread: function question

  1. #1
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    function question

    find f'''(x) if f(x)=xsinx

    is it just simply f'''(x)=-xcosx
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  2. #2
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    no you need to use the product rule...if f and g are functions, then the derivative of $\displaystyle f\cdot g=f'\cdot g+f\cdot g'$

    Use that rule and try again
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  3. #3
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    so f'(x)= sinx+xcosx
    f''(x)=cosx+(cosx-xsinx)
    f'''(x)=-sinx-sinx-cosx
    i dont think i did that right. can anyone help me. calc is kicking me hard
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  4. #4
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    $\displaystyle f(x)=x\sin x$

    $\displaystyle f'(x)= (x)'\cdot \sin x+x\cdot (\sin x)'$

    $\displaystyle =1\cdot \sin x+x\cdot \cos x$

    $\displaystyle =\sin x+x\cos x$

    $\displaystyle f''(x)=(\sin x)'+(x)'\cos x+x\cdot (\cos x)'$

    $\displaystyle =\cos x+\cos x-x\sin x=2\cos x-x\sin x$

    $\displaystyle f'''(x)=(2\cos x)'+(-x)'\sin x +(-x)(\sin x)'$

    $\displaystyle =-2\sin x-\sin x-x\cos x=-3\sin x-x\cos x$
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  5. #5
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    thank you very much.
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