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F(x,y) = sqrt2x^2+2y^2
fx(2,-3) =
In the photo too

F(x,y) = sqrt2x^2+2y^2 fx(2,-3) = In the photo too-example-1
User Risk
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1 Answer

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Final answer:

To find fx(2, -3), we differentiate the function F(x, y) = √(2x^2 + 2y^2) with respect to x and evaluate at the point (2, -3), resulting in fx(2, -3) = 4/√(26).

Step-by-step explanation:

The question involves calculating the partial derivative of the function F(x,y) with respect to x, evaluated at the point (2, -3).

To find fx(2, -3), we need to differentiate the function F(x,y) partially with respect to x while treating y as a constant and then plug in the values for x and y.

The given function is F(x,y) = √(2x^2 + 2y^2). The partial derivative with respect to x is:

fx(x,y) = ∂/∂x (√(2x^2 + 2y^2))
= 1/2(2x^2 + 2y^2)^{-1/2}*4x
= 2x/√(2x^2 + 2y^2)

Substituting x = 2 and y = -3 into the derivative, we get:

fx(2,-3) = 2*2/√(2*2^2 + 2*(-3)^2)
= 4/√(8 + 18)
= 4/√(26)
= 4/√(2*13)
= 4/√(26)

User Mehar
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