Question

find the partial derivative.
f(x,y)=8x + 4x^{2}y^{2}-3y^{2}. find f_{x}(x,y).
a. 8x^{2}y - 6y
b. 8 + 8xy^{2}
c. 8 + 8x^{2}y
d. 8xy^{2}-6y

Explanation:

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

• For the term \(8x\), using the power rule \(\frac{d}{dx}(ax^n)=anx^{n - 1}\) (here \(n = 1\)), \(\frac{\partial(8x)}{\partial x}=8\).
• For the term \(4x^{2}y^{2}\), using the power rule \(\frac{\partial(4x^{2}y^{2})}{\partial x}=4y^{2}\frac{\partial(x^{2})}{\partial x}\). Since \(\frac{\partial(x^{2})}{\partial x}=2x\), then \(4y^{2}\times2x = 8xy^{2}\).
• For the term \(-3y^{2}\), since \(y\) is treated as a constant when differentiating with respect to \(x\), \(\frac{\partial(-3y^{2})}{\partial x}=0\).

Step2: Sum up the derivatives of each term

\(f_{x}(x,y)=\frac{\partial(8x)}{\partial x}+\frac{\partial(4x^{2}y^{2})}{\partial x}+\frac{\partial(-3y^{2})}{\partial x}=8 + 8xy^{2}+0\)

Answer:

B. \(8 + 8xy^{2}\)