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Find ∂z/∂x and ∂z/∂y for x² + 2y² + 7z² = 1.

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Final answer:

For the equation x² + 2y² + 7z² = 1, ∂z/∂x is -x/(7z) and ∂z/∂y is -2y/(7z) after applying partial differentiation with respect to x and y, respectively.

Step-by-step explanation:

To find ∂z/∂x and ∂z/∂y for the given equation x² + 2y² + 7z² = 1, we need to use partial differentiation. Partial differentiation involves differentiating with respect to one variable while keeping the other variables constant.

For ∂z/∂x, differentiate the equation with respect to x, treating y and z as constants:

2x + 0 + 0 = 0 (since the derivative of y and z terms with respect to x are zero)

∂z/∂x = -2x/(14z) (after rearranging and solving for ∂z/∂x)

For ∂z/∂y, differentiate the equation with respect to y, treating x and z as constants:

0 + 4y + 0 = 0 (since the derivative of x and z terms with respect to y are zero)

∂z/∂y = -4y/(14z) (after rearranging and solving for ∂z/∂y)

To find ∂z/∂x and ∂z/∂y for the equation x² + 2y² + 7z² = 1, we need to differentiate the equation with respect to x and y.

When we differentiate x² + 2y² + 7z² = 1 with respect to x, we get 2x + 0 + 0 = 0. Therefore, ∂z/∂x = 0.

When we differentiate x² + 2y² + 7z² = 1 with respect to y, we get 0 + 4y + 0 = 0. Therefore, ∂z/∂y = 0.

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