Question

find the partial derivative.
f(x,y)=2x + 3x^{2}y^{2}-3y^{2}. find f_{x}(x,y).
a. 2 + 6x^{2}y
b. 6x^{2}y - 6y
c. 6xy^{2}-6y
d. 2 + 6xy^{2}

Explanation:

Step1: Differentiate each term with respect to \(x\)

When differentiating \(2x\) with respect to \(x\), using the power rule \(\frac{d}{dx}(ax^n)=anx^{n - 1}\) (here \(a = 2\), \(n=1\)), we get \(\frac{\partial}{\partial x}(2x)=2\).
For the term \(3x^{2}y^{2}\), since \(y\) is treated as a constant, using the power rule \(\frac{\partial}{\partial x}(3x^{2}y^{2})=3y^{2}\frac{\partial}{\partial x}(x^{2})\). And \(\frac{\partial}{\partial x}(x^{2}) = 2x\), so \(3y^{2}\times2x=6xy^{2}\).
For the term \(-3y^{2}\), since \(y\) is treated as a constant and there is no \(x\) in the term, \(\frac{\partial}{\partial x}(-3y^{2}) = 0\).

Step2: Sum up the derivatives of each term

\(f_{x}(x,y)=\frac{\partial}{\partial x}(2x)+\frac{\partial}{\partial x}(3x^{2}y^{2})+\frac{\partial}{\partial x}(-3y^{2})\)
\(f_{x}(x,y)=2 + 6xy^{2}+0\)

Answer:

D. \(2 + 6xy^{2}\)