Solving a Partial Derivative Problem with Substitution

Derill03
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Find par(z)/par(t) at s=1, t=0
when z= ln(x+y), x=s+t, y=s-t

Not sure how to approach cause if i plug in s's and t's i get an answer of 0 because taking the partial with respect to t yields a zero. Can someone shed some light on how to correctly solve?

par(z)/par(t) = partial derivative of z with respect to t
 
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find \frac{\partial{z}}{\partial{x}} \frac{\partial{x}}{\partial{t}}} and \frac{\partial{z}}{\partial{y}} \frac{\partial{y}}{\partial{t}}
 
The chain rule for partial derivatives is
\frac{\partial z}{\partial t}= \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}
 
I just don't know how to deal with it in the form its in. The s and t are what are confusing me, can u give me some sort of an example?

The way i did it is substitute s and t for y and x so i get ln(2s) but when u take partial with respect to t you get 0? is this correct?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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