partial differentiation

**curlywurlysqurly** · May 25th 2008, 12:19 PM

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

**galactus** · May 25th 2008, 12:28 PM

Just use the product rule as with regular diff except treat x as a constant.

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

$f_{y}=\sqrt{x^{2}+y^{2}}+\frac{y^{2}}{\sqrt{x^{2}+ y^{2}}}$

**curlywurlysqurly** · May 25th 2008, 12:42 PM

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

**Chris L T521** · May 25th 2008, 07:15 PM

Originally Posted by curlywurlysqurly

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

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

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

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

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

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

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

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

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

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

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

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