To find [tex]\( f_x(4,2) \)[/tex] for the function [tex]\( f(x, y) = 5 x^2 y - 2 x y^3 \)[/tex], we need to follow these steps:
1. Differentiate the function [tex]\( f(x, y) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
f(x, y) = 5 x^2 y - 2 x y^3
\][/tex]
Taking the partial derivative with respect to [tex]\( x \)[/tex]:
- For the term [tex]\( 5 x^2 y \)[/tex], use the power rule:
[tex]\[
\frac{\partial}{\partial x} (5 x^2 y) = 10 x y
\][/tex]
- For the term [tex]\( 2 x y^3 \)[/tex], use the product rule:
[tex]\[
\frac{\partial}{\partial x} (2 x y^3) = 2 y^3
\][/tex]
Therefore, the partial derivative with respect to [tex]\( x \)[/tex] is:
[tex]\[
f_x(x, y) = 10 x y - 2 y^3
\][/tex]
2. Evaluate the partial derivative at the point [tex]\( (4, 2) \)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = 2 \)[/tex] into [tex]\( f_x(x, y) \)[/tex]:
[tex]\[
f_x(4, 2) = 10 (4) (2) - 2 (2)^3
\][/tex]
Simplify the expression step-by-step:
- Calculate [tex]\( 10 \cdot 4 \cdot 2 \)[/tex]:
[tex]\[
10 \cdot 4 \cdot 2 = 80
\][/tex]
- Calculate [tex]\( 2 \cdot 2^3 \)[/tex]:
[tex]\[
2 \cdot 2^3 = 2 \cdot 8 = 16
\][/tex]
- Subtract the two results:
[tex]\[
80 - 16 = 64
\][/tex]
So, the value of [tex]\( f_x(4, 2) \)[/tex] is [tex]\( 64 \)[/tex].
