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Question

If ex+ey=ex+y, prove that dydx+ey-x=0

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Solution

We have, ex+ey=ex+y ...(1)
Differentiating both sides using chain rule,
ddxex+ddxey=ddxex+yex+eydydx=ex+yddxx+yex+eydydx=ex+y1+dydxeydydx-ex+ydydx=ex+y-exdydx=ex+y-exey-ex+ydydx=ex+y-ex+y+eyex+y-ex-ex+y Using eqn.1dydx=ey-exdydx=-ey-xdydx+ey-x=0

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