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Find the first partial derivatives of f(x,y)=3x-4y3x 4y at the point (x,y)=(1,3).

User Rochele
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Final answer:

The first partial derivatives of the function f(x,y)=3x-4y^3 at the point (1,3) are 3 and -108 for x and y respectively.

Step-by-step explanation:

The student asked to find the first partial derivatives of the function f(x,y)=3x-4y3 at the point (x,y)=(1,3). We start by computing the partial derivatives with respect to x and y. For ∂f/∂x (partial derivative with respect to x), we treat y as constant and differentiate f with respect to x which simply results in the constant 3 since y has no x to interact with. For ∂f/∂y (partial derivative with respect to y), we treat x as constant and differentiate -4y3 with respect to y which gives us -12y2. Now, we evaluate these partial derivatives at the point (1,3). Therefore, ∂f/∂x = 3 and ∂f/∂y at (1,3) is -12*32 which equals -108. The first partial derivatives of the function at the given point are 3 and -108.

User Apnorton
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