Find the first partial derivative of the functon
f
(
x
,
y
)
=
2
x
3
y
2
+
y
3
assuming you meant
f(x,y)=2x^3y^2+y^3
fx = 6x^2y^2
fy = 4x^3y+3y^2
To find the first partial derivative of the function f(x, y) = 2x^3y^2 + y^3 with respect to x, we differentiate the function with respect to x while treating y as a constant.
The partial derivative of f(x, y) with respect to x is denoted as ∂f/∂x or df/dx. To find it, we differentiate the function with respect to x, treating y as a constant. The derivative of 2x^3y^2 with respect to x is obtained by applying the power rule for differentiation.
Let's go step by step:
1. Differentiate 2x^3y^2 with respect to x:
- The derivative of 2x^3 with respect to x is 6x^2.
- Since y^2 is treated as a constant, it remains unchanged.
2. Differentiate y^3 with respect to x:
- The derivative of y^3 with respect to x is 0, as y is treated as a constant.
Combining the results of the two differentiations, the first partial derivative of f(x, y) = 2x^3y^2 + y^3 with respect to x is:
∂f/∂x = 6x^2y^2
So, the first partial derivative of f(x, y) with respect to x is 6x^2y^2