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Math Help - Implicit Differentiation of a Trignometric Function

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Implicit Differentiation of a Trignometric Function

    Use implicit differentiation to find the second derivative:

    > y+ sin y = x

    dy/dx + cosy dy/dx = 1

    dydx (1+cosy) = 1

    dy/dx = 1/1+cosy

    d^2y/dx^2 = /frac{1+cosy*0 - 1*siny*dy/dx}{(1+cosy)^2}

    \frac{0-siny*dy/dx}{(1+cosy)^2}

    But the answer is

    \frac{siny}{(1+cosy)^3}

    What did I do wrong ?
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  2. #2
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    Quote Originally Posted by StarlitxSunshine View Post
    Use implicit differentiation to find the second derivative:

    > y+ sin y = x

    dy/dx + cosy dy/dx = 1

    dy/dx (1+cosy) = 1

    dy/dx = 1/1+cosy


    Up to this place all's right, but now you have to differentiate the right side, and get \frac{d^2y}{dx^2}=\frac{\sin x}{(1+\cos x)^2}.
    I can't see from where that 3 in the power of the denominator in what you call "the answer" comes...

    Tonio

    d^2y/dx^2 = /frac{1+cosy*0 - 1*siny*dy/dx}{(1+cosy)^2}

    \frac{0-siny*dy/dx}{(1+cosy)^2}

    But the answer is

    \frac{siny}{(1+cosy)^3}

    What did I do wrong ?

    .
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  3. #3
    Junior Member StarlitxSunshine's Avatar
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    My textbook has answers to all the odd numbered questions. The denominator is definitely cubed.

    & with the differentiation, I'm still getting -sin y not sin y.
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    Quote Originally Posted by StarlitxSunshine View Post
    My textbook has answers to all the odd numbered questions. The denominator is definitely cubed.

    & with the differentiation, I'm still getting -sin y not sin y.
    It is a quotient derivative: \left(\frac{u}{v}\right)'=\frac{u'v-v'u}{v^2}, and since the derivative of cos x is -sin x , with the sign minus it must be plus...and squared in the denominator.
    What book do you have?

    Tonio
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  5. #5
    Junior Member StarlitxSunshine's Avatar
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    Oh! I see where I went wrong with the minus sign Oo Thank you =)

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