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Question
Find `"dy"/"dx"` if xey + yex = 1
Solution
xey + yex = 1
Differentiating both sides w.r.t. x, we get
`"d"/"dx"(xe^y) + "d"/"dx"(ye^x)` = 0
∴ `x."d"/"dx"(e^y) + e^y."d"/"dx"(x) + y."d"/"dx"(e^x) + e^x."dy"/"dx"` = 0
∴ `x.e^y"dy"/"dx" + e^y xx 1 + y xx e^x + e^x"dy"/"dx"`= 0
∴ `(e^x + xe^y)"dy"/"dx"` = – ey – yex
∴ `"dy"/"dx" = -((e^y + ye^x)/(e^x + xe^y))`.
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