# Partial differentiation of integrals from u(x, y) to v(x, y)

I'm having trouble with this question. If: $$w = \displaystyle\int_{xy}^{2x-3y}du/ln(u)\,du$$

Find $$\frac{\partial y}{\partial x}$$ at x = 3, y= 1.

I know that the general rule for differentiation of integrals of this type goes:

$$\frac{d}{dx}\displaystyle\int_{u(x)}^{v(x)}f(t)dt = f(v)\frac{dv}{dx} - f(u)\frac{du}{dx}$$

Using the general rule and substituting partial derivatives for singular derivatives, I found: $$\frac{\partial w}{\partial x} = 1/ln(3)$$ and $$\frac{\partial w}{\partial y} = -6/ln(3)$$

However, I am at a loss as to how to find the solution to this question as the partial derivative is of the terms that define the integral's interval.

Would I be able to consider the equation:

$$\frac{\partial y}{\partial x} = \frac{\partial y}{\partial w}\frac{\partial w}{\partial x}$$

and then say $$\frac{\partial y}{\partial w} = 1/\frac{\partial w}{\partial y}$$

Any help would be greatly appreciated.

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Is $y$ a function in $x$? –  Mhenni Benghorbal Mar 5 '13 at 0:04
I didn't quite get what's under the integral. –  Kaster Mar 5 '13 at 0:04

This is the sort of confusion fueled by our suboptimal notation for partial derivatives, which fails to indicate which variables are being held constant.

I'll assume that you're considering all of $w,x,y$ as functions of the other two, so when you write $\def\p#1#2{\dfrac{\partial #1}{\partial #2}}\p wx$ you mean $\def\pp#1#2#3{\left.\dfrac{\partial #1}{\partial #2}\right|_{#3}}\pp wxy$ and when you write $\displaystyle\p yx$ you mean $\pp yxw$ (where the variable indicated underneath the vertical bar is being held constant).

Under this interpretation, your second equation is correct,

$$\pp ywx=\left(\pp wyx\right)^{-1}\;,$$

which just expresses $\def\pd#1#2{\dfrac{\mathrm d#1}{\mathrm d#2}}\pd yw=\left(\pd wy\right)^{-1}$ for the univariate functions you get when you treat $x$ as a fixed parameter. However, your first equation generally doesn't hold, that is, generally

$$\pp yxw\ne\pp ywx\pp wxy\;.$$

Informally speaking, you can't cancel $\partial w$ because different variables are being held constant in the different derivatives (in contrast to the second equation).

To find $\pp yxw$ you can apply the chain rule:

$$0=\pp wxw=\pp wxy\pp xxw+\pp wyx\pp yxw=\pp wxy+\pp wyx\pp yxw\;,$$

and thus

$$\pp yxw=-\frac{\pp wxy}{\pp wyx}=\frac16\;.$$

Another way to look at this is to regard $w$ as constant and set the differential $\mathrm dw$ to zero:

$$\mathrm dw=\pp wxy\mathrm dx+\pp wyx\mathrm dy=0\;,$$

so

$$\pd yx=-\frac{\pp wxy}{\pp wyx}\;,$$

and $\pd yx$ with $w$ held constant is $\pp yxw$.

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I think notation like $$\left .\frac {\partial w}{\partial y}\right |_x^{-1}$$ where it means reciprocal of $\frac {\partial w}{\partial y}$ even more confusing, since it looks like $$\left .\frac {\partial w}{\partial y}\right |_{-1} - \left . \frac {\partial w}{\partial y} \right|_x$$ I'd suggest to use $$\left . \frac {\partial y}{\partial w}\right |_x = \left (\left . \frac {\partial w}{\partial y}\right |_x \right )^{-1}$$ –  Kaster Mar 5 '13 at 0:50
Thanks Joriki, your explanation makes sense. I think my biggest hurdle to solving this problem was that I did not realize that for $$\frac{\partial y}{\partial x}$$ I am allowed to treat w as a constant.I can see how the notation, had it been satisfactory, would have helped me realize my mistake. –  user65019 Mar 5 '13 at 1:58
@Kaster: I fully agree and I've fixed that. In fact I had it like that originally, then felt that it looked too cluttered with the parentheses, and then didn't realize when I removed them how ambiguous that makes the expression. Thanks for pointing that out. –  joriki Mar 5 '13 at 5:14