Partial derivative chain rule question

engineer_dave
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Homework Statement



Given z= square root of xy, x = 2t - 1, y = 3t +4, use the chain rule to find dz/dt as a function of t.

Homework Equations





The Attempt at a Solution



dz/dt = partial derivative of z with respect to x multiplied by dx/dt + (partial derivative of z with respect to y multiplied by dy/dt)

I got that part right but how do you differentiate square root of xy as a partial derivative of z with respect to x?? Can you show me the final answer of that?

Thanks!
 
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take y as a constant and find the partial derivative of z=sqrt(xy) like you normally would.
 
yea i tried that but could u give me the final answer to that particular part. would it 1/2x^-1/2 multiplied by y?
 
you're missing a y with the x, it should be \frac{(xy)'}{2\sqrt{xy}}

where (xy)'=y

when you take the partial derivative you're just leaving y as constant so the y stays with x under the square root.
 
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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