QuestionJuly 14, 2025

Given f(x,y)=3x^3-3xy^2-5y^4 find the following numerical values: f_(x)(4,4)=square f_(y)(4,4)=square

Given f(x,y)=3x^3-3xy^2-5y^4 find the following numerical values: f_(x)(4,4)=square f_(y)(4,4)=square
Given f(x,y)=3x^3-3xy^2-5y^4 find the following numerical values: f_(x)(4,4)=square f_(y)(4,4)=square

Solution
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Answer

f_{x}(4,4)=96 ### f_{y}(4,4)=-1376 Explanation 1. Find the partial derivative with respect to x Differentiate f(x,y)=3x^{3}-3xy^{2}-5y^{4} with respect to x: f_{x}(x,y) = \frac{\partial}{\partial x}(3x^3 - 3xy^2 - 5y^4) = 9x^2 - 3y^2. 2. Evaluate f_{x}(4,4) Substitute x=4 and y=4 into f_{x}(x,y): f_{x}(4,4) = 9(4)^2 - 3(4)^2 = 144 - 48 = 96. 3. Find the partial derivative with respect to y Differentiate f(x,y)=3x^{3}-3xy^{2}-5y^{4} with respect to y: f_{y}(x,y) = \frac{\partial}{\partial y}(3x^3 - 3xy^2 - 5y^4) = -6xy - 20y^3. 4. Evaluate f_{y}(4,4) Substitute x=4 and y=4 into f_{y}(x,y): f_{y}(4,4) = -6(4)(4) - 20(4)^3 = -96 - 1280 = -1376.

Explanation

1. Find the partial derivative with respect to $x$<br /> Differentiate $f(x,y)=3x^{3}-3xy^{2}-5y^{4}$ with respect to $x$: $f_{x}(x,y) = \frac{\partial}{\partial x}(3x^3 - 3xy^2 - 5y^4) = 9x^2 - 3y^2$.<br />2. Evaluate $f_{x}(4,4)$<br /> Substitute $x=4$ and $y=4$ into $f_{x}(x,y)$: $f_{x}(4,4) = 9(4)^2 - 3(4)^2 = 144 - 48 = 96$.<br />3. Find the partial derivative with respect to $y$<br /> Differentiate $f(x,y)=3x^{3}-3xy^{2}-5y^{4}$ with respect to $y$: $f_{y}(x,y) = \frac{\partial}{\partial y}(3x^3 - 3xy^2 - 5y^4) = -6xy - 20y^3$.<br />4. Evaluate $f_{y}(4,4)$<br /> Substitute $x=4$ and $y=4$ into $f_{y}(x,y)$: $f_{y}(4,4) = -6(4)(4) - 20(4)^3 = -96 - 1280 = -1376$.
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