Question

Find `"dy"/"dx"` if xe^y + ye^x = 1

Answer 1

xe^y + ye^x = 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"` = – e^y – ye^x 

∴ `"dy"/"dx" = -((e^y + ye^x)/(e^x + xe^y))`.