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Math Help - Derivative of a Function at a point

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Derivative of a Function at a point

    If xsiny +ysinx = 0, then, \frac{dy}{dx} at \frac{\pi}{4},\frac{\pi}{4} = ?

    1. (A) -1
    2. (B) 1
    3. (C) 0
    4. (D) \frac{\pi}{4}
    5. (E)  \frac{-\pi}{4}


    I think I made a mistake somewhere in my working. Here's what I did:

    Using implicit differentiation:

    <br />
\frac{dy}{dx}[xsiny +ysinx] = \frac{dy}{dx} [0]<br />

    <br />
xcosy + siny +ycosx + sinyy'  0<br />

    <br />
sinyy' = -ycosx-siny-xcosy<br />

    <br />
y' = \frac{-ycosx-siny-xcosy}{siny}<br />

    <br />
y' = \frac{\frac{-\Pi}{4}cos\frac{-\Pi}{4} - sin\frac{-\Pi}{4} -\frac{-\pi}{4}cos\frac{-\Pi}{4}}{sin\frac{-\Pi}{4}}<br />

    <br />
y' = \frac{ -45cos45-sin45-45cos45}{sin45}<br />

    <br />
y' = \frac{-45\frac{\sqrt{2}} - \frac{\sqrt{2}}{2} - 45\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}<br />

    <br />
y' = -91\frac{\sqrt[2]{2}}{2}*\frac{2}{\sqrt{2}}<br />

    <br />
y'= -91
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  2. #2
    Super Member
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    I think you did the implicit differentiation wrong. Here's what I obtained:

    \frac{d}{dx}[xsin(y)+ysin(x)]=\frac{dx}{dx}sin(y)+x\frac{d}{dx}sin(y)+\frac{dy}  {dx}sin(x)+y\frac{d}{dx}sin(x)

    Substitute these values in the expression above:

    \frac{d}{dx}sin(x)=cos(x)

    \frac{d}{dx}sin(y)=\frac{dy}{dx}cos(y)

    So now we have:

    sin(y)+x\frac{dy}{dx}cos(y)+\frac{dy}{dx}sin(x)+yc  os(x)=0

    Try solving for \frac{dy}{dx} and see if you can get an answer.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by StarlitxSunshine View Post
    If xsiny +ysinx = 0, then, \frac{dy}{dx} at \frac{\pi}{4},\frac{\pi}{4} = ?

    1. (A) -1
    2. (B) 1
    3. (C) 0
    4. (D) \frac{\pi}{4}
    5. (E)  \frac{-\pi}{4}
    I think I made a mistake somewhere in my working. Here's what I did:

    Using implicit differentiation:

    <br />
\frac{dy}{dx}[xsiny +ysinx] = \frac{dy}{dx} [0]<br />

    <br />
xcosy + siny +ycosx + sinyy' 0<br />

    <br />
sinyy' = -ycosx-siny-xcosy<br />

    <br />
y' = \frac{-ycosx-siny-xcosy}{siny}<br />

    <br />
y' = \frac{\frac{-\Pi}{4}cos\frac{-\Pi}{4} - sin\frac{-\Pi}{4} -\frac{-\pi}{4}cos\frac{-\Pi}{4}}{sin\frac{-\Pi}{4}}<br />

    <br />
y' = \frac{ -45cos45-sin45-45cos45}{sin45}<br />

    <br />
y' = \frac{-45\frac{\sqrt{2}} - \frac{\sqrt{2}}{2} - 45\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}<br />

    <br />
y' = -91\frac{\sqrt[2]{2}}{2}*\frac{2}{\sqrt{2}}<br />

    <br />
y'= -91
    You must understand that y is a function of x. Therefore whenevever you take the derivative of y, treat is as such...

    (xcosy\frac{dy}{dx}+siny)+(ycosx+\frac{dy}{dx}siny  )=0
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  4. #4
    Junior Member StarlitxSunshine's Avatar
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    Yup Yup :3

    Then the answer is -1

    && I see what I did wrong, too. I didn't take the derivative properly when I did product rule :33

    Thank you very much ! =)
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