Calculate Partial Derivatives f_x(0,3) and f_y(0,3) for f(x, y) = -5x^2 + 3xy^3 + 6y^4

Math Problem Statement

Given f(x, y) = -5x^2 + 3xy^3 + 6y^4, find the following numerical values: f_x(0, 3) = f_y(0, 3) =

We are tasked with finding the partial derivatives $f_x(0,3)$ and $f_y(0,3)$ of the given function:

$f(x, y) = -5x^2 + 3xy^3 + 6y^4$

= -10x + 3y^3$$ Now substitute $$x = 0$$ and $$y = 3$$: $$f_x(0, 3) = -10(0) + 3(3)^3 = 3 \cdot 27 = 81$$ Thus, $$f_x(0, 3) = 81$$. --- ### Step 2: Find the partial derivative with respect to $$y$$, $$f_y(x, y)$$ - Differentiate $$f(x, y)$$ with respect to $$y$$ (treat $$x$$ as a constant): $$f_y(x, y) = \frac{\partial}{\partial y} \left(-5x^2 + 3xy^3 + 6y^4\right) = 0 + 9xy^2 + 24y^3$$ Now substitute $$x = 0$$ and $$y = 3$$: $$f_y(0, 3) = 9(0)(3)^2 + 24(3)^3 = 0 + 24(27) = 648$$ Thus, $$f_y(0, 3) = 648$$. --- ### Final Results: $$f_x(0, 3) = 81, \quad f_y(0, 3) = 648$$ Would you like a detailed explanation of partial derivatives or more examples to practice? --- ### Expanding Your Understanding: 1. What is the geometric interpretation of a partial derivative in this context? 2. How do partial derivatives help identify critical points in multivariable calculus? 3. What are second-order partial derivatives, and how are they computed? 4. How can mixed partial derivatives ($$f_{xy}$$ or $$f_{yx}$$) be evaluated for this function? 5. Can you generalize this result for arbitrary $$x$$ and $$y$$ instead of specific values? **Tip**: Always double-check substitution steps when evaluating partial derivatives to avoid errors in simplifying terms.

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Calculus
Partial Derivatives
Multivariable Functions

f_x(x, y) = ∂f/∂x
f_y(x, y) = ∂f/∂y

Partial Differentiation

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