Hi guys could you please help me with this question- For an arbitrary function f = f(u, v) show that 1/x(df/dx) + 1/y(df/dy) = 4(df/du) where u = x^2 + y^2, v = x^2 − y^2, x not = 0 and y not = 0
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Originally Posted by rologolfer Hi guys could you please help me with this question- For an arbitrary function f = f(u, v) show that 1/x(df/dx) + 1/y(df/dy) = 4(df/du) where u = x^2 + y^2, v = x^2 − y^2, x not = 0 and y not = 0 By the chain rule, $\displaystyle \frac{\partial f}{\partial x}= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}$. Does this help?
Thanks a lot, thats great!
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