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Question

Find dydx when y=ex+y

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Solution

y=ex+y
Taking log
log(y)=(x+y)logee logee=1
Differentiating w.r.t. x
1y(dydx)=1+dydx
(dydx)(1y1)=1
(dydx)=y1y=ex+y1ex+y.

1230489_1301587_ans_d52f0b4233d846be8aac3b1d47544f1e.jpg

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