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If u=log (x^2 y^2) then x du/dx y du/dy=.?
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If u=log (x^2 y^2) then x du/dx y du/dy=.?



Explanation:

Given: u = log(x^2 y^2)
- We need to find x * du/dx + y * du/dy

Chain Rule:
- The chain rule states that if u = f(g(x)), then du/dx = f'(g(x)) * g'(x)

Applying Chain Rule:
- Here, u = log(x^2 y^2)
- Let's consider f(z) = log(z) where z = x^2 y^2
- f'(z) = 1/z

Partial Derivatives:
- To find du/dx and du/dy, we need to take partial derivatives
- du/dx = f'(z) * d/dx (x^2 y^2)
= 1/(x^2 y^2) * 2x * y^2
= 2y/x
- du/dy = f'(z) * d/dy (x^2 y^2)
= 1/(x^2 y^2) * 2y * x^2
= 2x/y

Final Calculation:
- x * du/dx + y * du/dy
= x * (2y/x) + y * (2x/y)
= 2y + 2x
= 2(x + y)

Conclusion:
- Therefore, x * du/dx + y * du/dy = 2(x + y)
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If u=log (x^2 y^2) then x du/dx y du/dy=.?
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