Find the Four 2nd Partial Derivatives of f(x,y)

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SUMMARY

The discussion focuses on finding the four second partial derivatives of the function f(x,y) = y²e^x + xycos(x). The user successfully calculated Zxx = y²e^x - ycos(x) and Zy = 2ye^x + xcos(x), but required assistance with Zxy and Zyx. The key takeaway is the importance of treating the non-differentiated variable as a constant during differentiation, which is crucial for obtaining accurate results in higher derivatives.

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  • Understanding of partial derivatives
  • Familiarity with the exponential function and trigonometric functions
  • Knowledge of the product rule in differentiation
  • Basic calculus concepts, particularly in multivariable calculus
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  • Practice finding second partial derivatives of multivariable functions
  • Study the application of the mixed partial derivatives theorem
  • Explore the implications of Clairaut's theorem on the equality of mixed partial derivatives
  • Review the product rule and chain rule in the context of multivariable calculus
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Find the four second partial derivatives for f(x,y) = y^2e^x + xycosx

I am stuck on the last part... here's what I got so far:

Zx = y^2e^x - ysinx
Zxx = y^2e^x - ycosx
Zxy = ?

Zy = 2ye^x + xcosx
Zyy = 2e^x + xcosx
Zyx = ?


I need help with solving for xy. Both should end up being the same. :eek:
 
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the trick is to ignore the other variable whenever you are not differentiating with respect to it. eg.
\frac{\partial}{\partial x} x e^y = e^y and
\frac{\partial}{\partial y} x e^y = x e^y

in first case you treat y as if it is a constant whereas in the second you treat x as constant instead. the same rules apply when you go to higher derviatives
eg.
\frac{\partial}{\partial x} \frac{\partial}{\partial y} (x e^y) <br /> = \frac{\partial}{\partial x}(x e^y) = e^y
 
by the way your Zx is wrong
 

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