RazerM
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Homework Statement
Find \frac{\partial z}{\partial x} \frac{\partial z}{\partial y} where z=\left( [x+y]^3-4y^2 \right)^{\frac{1}{2}}
Homework Equations
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The Attempt at a Solution
I know that \frac{\partial z}{\partial y}=\frac{3(x+y)^2-8y}{2\sqrt{(x+y)^3-4y^2}}
but I am unsure whether \frac{\partial z}{\partial x} is the exact same or does not include the '-8y' in the numerator.
I get the feeling that when finding the derivative inside the square root (in z) that y should still be treated as constant and therefore have no -8y.
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