Find the partial derivative of f(x, y) = x3 + 2x3y3 – 3y2 + x + y, at (2,1)

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Find the partial derivative of f(x, y) = x3 + 2x3y3 &#8211; 3y2 + x + y, at (2,1)

Suresh Chand · Accepted Answer

Solution: (f(x, y) = x^3 + 2x^{3}y^{3} - 3y^2 + x + y) First we wish to find (frac{∂f}{∂x}). For this, we suppose y as constant value and differentiate f with respect to x, we get,  (frac{∂f}{∂x} = frac{∂}{∂x}(x^3 + 2x^{3}y^{3} - 3y^2 + x + y)) = 3x2 + 6x2y3 + 1 At point (2, 1), the value of (frac{∂f}{∂x}) is, (frac{∂f}{∂x}left |_{(2, 1)} right. ) (= 3times8 + 6times4times1 + 1 = 49) Next, we wish  to find (frac{∂f}{∂y}). For this, we suppose x as constant value and differentiate f with respect to y, we get, (frac{∂f}{∂x}) ( = frac{∂(x^3 + 2x^{3}y^{3} - 3y^2 + x + y)}{∂y}) = 6x3y2 - 6y + 1 At point (2, 1), the value of (frac{∂f}{∂y}) is, (frac{∂f}{∂y}left |_{(2, 1)} right. ) = 6 × 8 × 1 - 6 + 1 = 43 Thus fx(2, 1) = 49 and fy(2, 1) = 43

Find the partial derivative of f(x, y) = x³ + 2x³y³ – 3y² + x + y, at (2,1)

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Find the partial derivative of f(x, y) = x3 + 2x3y3 – 3y2 + x + y, at (2,1)

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Find the partial derivative of f(x, y) = x³ + 2x³y³ – 3y² + x + y, at (2,1)