Solving Partial Derivative of Z with respect to x

The problem asks for the partial derivative of the function $Z = 2x^{2}y^{2} + 3xy^{2} + y^{3}$ with respect to $x$. Remember that when finding a partial derivative, you treat all variables other than the one you're differentiating with respect to as constants.

Calculating $\frac{\partial Z}{\partial x}$

We differentiate each term of $Z$ individually with respect to $x$, keeping $y$ constant:

• Derivative of $2x^{2}y^{2}$: Treat $y^{2}$ as a constant. The derivative of $2x^{2}$ with respect to $x$ is $2 \cdot 2x = 4x$. So, the derivative of this term is $4xy^{2}$. $ \frac{\partial}{\partial x}(2x^{2}y^{2}) = 2y^{2} \frac{\partial}{\partial x}(x^{2}) = 2y^{2}(2x) = 4xy^{2} $
• Derivative of $3xy^{2}$: Treat $y^{2}$ as a constant. The derivative of $3x$ with respect to $x$ is $3$. So, the derivative of this term is $3y^{2}$. $ \frac{\partial}{\partial x}(3xy^{2}) = 3y^{2} \frac{\partial}{\partial x}(x) = 3y^{2}(1) = 3y^{2} $
• Derivative of $y^{3}$: Since $y^{3}$ does not contain $x$, it is treated as a constant. The derivative of any constant with respect to $x$ is $0$. $ \frac{\partial}{\partial x}(y^{3}) = 0 $

Final Partial Derivative

Combine the results from differentiating each term:

$ \frac{\partial Z}{\partial x} = 4xy^{2} + 3y^{2} + 0 $

Therefore, the partial derivative of $Z$ with respect to $x$ is $4xy^{2} + 3y^{2}$.