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If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0?
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If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0?
Proof:
To prove that du/dx * du/dy * dy/dz = 0, we need to find the partial derivatives of u with respect to x, y, and z and then multiply them together.

Step 1: Find the partial derivative of u with respect to x (du/dx)

Taking the partial derivative of u with respect to x involves treating y and z as constants and differentiating u with respect to x.

Given u = x²(y-z)(y²(z-x))(z²(x-y))

To differentiate u with respect to x, we apply the product rule and chain rule:

- Using the product rule, we differentiate each term separately and add them together.
- For the first term, x², the derivative with respect to x is 2x.
- For the second term, (y-z), the derivative with respect to x is 0 because y and z are constants.
- For the third term, (y²(z-x)), the derivative with respect to x is -y².
- For the fourth term, (z²(x-y)), the derivative with respect to x is z².

Therefore, du/dx = 2x(y-z)(y²(z-x))(z²(x-y)) - y²(y-z)(z²(x-y)) + z²(y-z)(y²(z-x))

Step 2: Find the partial derivative of u with respect to y (du/dy)

Taking the partial derivative of u with respect to y involves treating x and z as constants and differentiating u with respect to y.

Given u = x²(y-z)(y²(z-x))(z²(x-y))

To differentiate u with respect to y, we again apply the product rule and chain rule:

- Using the product rule, we differentiate each term separately and add them together.
- For the first term, x², the derivative with respect to y is 0 because x is a constant.
- For the second term, (y-z), the derivative with respect to y is 1.
- For the third term, (y²(z-x)), the derivative with respect to y is 2y(z-x).
- For the fourth term, (z²(x-y)), the derivative with respect to y is 0 because x and z are constants.

Therefore, du/dy = x² + (y-z)(2y(z-x))(z²(x-y))

Step 3: Find the partial derivative of y with respect to z (dy/dz)

Taking the partial derivative of y with respect to z involves treating x as a constant and differentiating y with respect to z.

Given y = x²(y-z)(y²(z-x))(z²(x-y))

To differentiate y with respect to z, we apply the product rule and chain rule:

- Using the product rule, we differentiate each term separately and add them together.
- For the first term, x², the derivative with respect to z is 0 because x is a constant.
- For the second term, (y-z), the derivative with respect to z is -1.
- For the third term, (y²(z-x)), the derivative with respect to z is y².
- For the fourth term, (z²(x-y)), the derivative with
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If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0?
If u =x�(y-z)+y�(z-x)+z�(x-y) then prove that du/dx +du/dy +du/dz =0
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If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0?
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If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If u=x²(y-z) y²(z-x) z²(x-y) prove that du/dx du/dy dy/dz=0?.
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