find the partial derivative.\nf(x,y)=2x + 3x^{2}y^{2}-3y^{2}. find f_{x}(x,y).\na. 2 + 6x^{2}y\nb. 6x^{2}y…

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})