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Thread: 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 $\displaystyle xsiny +ysinx = 0$, then, $\displaystyle \frac{dy}{dx}$ at $\displaystyle \frac{\pi}{4},\frac{\pi}{4}$ = ?

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


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

    Using implicit differentiation:

    $\displaystyle
    \frac{dy}{dx}[xsiny +ysinx] = \frac{dy}{dx} [0]
    $

    $\displaystyle
    xcosy + siny +ycosx + sinyy' 0
    $

    $\displaystyle
    sinyy' = -ycosx-siny-xcosy
    $

    $\displaystyle
    y' = \frac{-ycosx-siny-xcosy}{siny}
    $

    $\displaystyle
    y' = \frac{\frac{-\Pi}{4}cos\frac{-\Pi}{4} - sin\frac{-\Pi}{4} -\frac{-\pi}{4}cos\frac{-\Pi}{4}}{sin\frac{-\Pi}{4}}
    $

    $\displaystyle
    y' = \frac{ -45cos45-sin45-45cos45}{sin45}
    $

    $\displaystyle
    y' = \frac{-45\frac{\sqrt{2}} - \frac{\sqrt{2}}{2} - 45\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}
    $

    $\displaystyle
    y' = -91\frac{\sqrt[2]{2}}{2}*\frac{2}{\sqrt{2}}
    $

    $\displaystyle
    y'= -91$
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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:

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

    Substitute these values in the expression above:

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

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

    So now we have:

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

    Try solving for $\displaystyle \frac{dy}{dx}$ 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 $\displaystyle xsiny +ysinx = 0$, then, $\displaystyle \frac{dy}{dx}$ at $\displaystyle \frac{\pi}{4},\frac{\pi}{4}$ = ?

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

    Using implicit differentiation:

    $\displaystyle
    \frac{dy}{dx}[xsiny +ysinx] = \frac{dy}{dx} [0]
    $

    $\displaystyle
    xcosy + siny +ycosx + sinyy' 0
    $

    $\displaystyle
    sinyy' = -ycosx-siny-xcosy
    $

    $\displaystyle
    y' = \frac{-ycosx-siny-xcosy}{siny}
    $

    $\displaystyle
    y' = \frac{\frac{-\Pi}{4}cos\frac{-\Pi}{4} - sin\frac{-\Pi}{4} -\frac{-\pi}{4}cos\frac{-\Pi}{4}}{sin\frac{-\Pi}{4}}
    $

    $\displaystyle
    y' = \frac{ -45cos45-sin45-45cos45}{sin45}
    $

    $\displaystyle
    y' = \frac{-45\frac{\sqrt{2}} - \frac{\sqrt{2}}{2} - 45\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}
    $

    $\displaystyle
    y' = -91\frac{\sqrt[2]{2}}{2}*\frac{2}{\sqrt{2}}
    $

    $\displaystyle
    y'= -91$
    You must understand that y is a function of x. Therefore whenevever you take the derivative of y, treat is as such...

    $\displaystyle (xcosy\frac{dy}{dx}+siny)+(ycosx+\frac{dy}{dx}siny )=0$
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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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