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Thread: partial differentiation

  1. #1
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    partial differentiation

    hi, im stuck trying to differentiate this with respect to y:

    y(x^2+y^2)^1/2

    ive tried the product rule and the chain rule and its gettin a bit complicated
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  2. #2
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    Just use the product rule as with regular diff except treat x as a constant.

    $\displaystyle f=y(\frac{1}{2})(x^{2}+y^{2})^{\frac{-1}{2}}(2y)+(x^{2}+y^{2})^{\frac{1}{2}}$

    $\displaystyle f_{y}=\sqrt{x^{2}+y^{2}}+\frac{y^{2}}{\sqrt{x^{2}+ y^{2}}}$
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  3. #3
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    oh right, well i got the same, only with the 1st term multiplied by y :s either way, its still a bit compicated. im actually doin a greens theorem problem

    integral {(xy - 4x^4y - y(x^2+y^2)^1/2 + 1/2 y^2sinx)dx + (4xy^4 + 8/3x^3y^2 + x(x^2+y^2)^1/2 - ycosx) dy}

    its not cancelling like it ought to when i try it
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by curlywurlysqurly View Post
    oh right, well i got the same, only with the 1st term multiplied by y :s either way, its still a bit compicated. im actually doin a greens theorem problem

    integral {(xy - 4x^4y - y(x^2+y^2)^1/2 + 1/2 y^2sinx)dx + (4xy^4 + 8/3x^3y^2 + x(x^2+y^2)^1/2 - ycosx) dy}

    its not cancelling like it ought to when i try it
    $\displaystyle \oint (xy-4x^4y-y\sqrt{x^2+y^2}+\frac{1}{2}y^2\sin x)dx+(4xy^4+\frac{8}{3}x^3y^2+x\sqrt{x^2+y^2}-y\cos x)dy$

    since $\displaystyle f(x)=xy-4x^4y-y\sqrt{x^2+y^2}+\frac{1}{2}y^2\sin x$, we find that $\displaystyle \frac{\partial f}{\partial y}=x-4x^4-\left[\frac{y^2}{\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}\right]+y\sin x$
    $\displaystyle \implies\frac{\partial f}{\partial y}=x-4x^4-\left[\frac{x^2+2y^2}{\sqrt{x^2+y^2}}\right]+y\sin x$

    since $\displaystyle g(x)=4xy^4+\frac{8}{3}x^3y^2+x\sqrt{x^2+y^2}-y\cos x$, we find that $\displaystyle \frac{\partial g}{\partial x}=4y^4+8x^2y^2+\left[\frac{x^2}{\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}\right]+y\sin x$
    $\displaystyle \implies \frac{\partial g}{\partial x}=4y^4+8x^2y^2+\left[\frac{2x^2+y^2}{\sqrt{x^2+y^2}}\right]+y\sin x$

    $\displaystyle \oint (xy-4x^4y-y\sqrt{x^2+y^2}+\frac{1}{2}y^2\sin x)dx+(4xy^4+\frac{8}{3}x^3y^2+x\sqrt{x^2+y^2}-y\cos x)dy$

    $\displaystyle \implies\iint_{R}\left[\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right]\,dA$$\displaystyle =\iint\left(4y^4+8x^2y^2+\frac{2x^2+y^2}{\sqrt{x^2 +y^2}}+y\sin x\right)-\left(x-4x^4-\frac{x^2+2y^2}{\sqrt{x^2+y^2}}+y\sin x\right)\,dA$

    $\displaystyle =\iint 4y^4+8x^2y^2-x+4x^4+\frac{2x^2+y^2}{\sqrt{x^2+y^2}}+\frac{2y^2+ x^2}{\sqrt{x^2+y^2}}\,dA$

    $\displaystyle =\iint 4y^4+8x^2y^2-x+4x^4+3\frac{x^2+y^2}{\sqrt{x^2+y^2}}\,dA$

    $\displaystyle =\iint 4y^4+8x^2y^2-x+4x^4+3\sqrt{x^2+y^2}\,dA$

    Use $\displaystyle C$ to find the limits for $\displaystyle R$...

    I hope this helped. (From what I can tell, only the trig terms canceled out)
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