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Question Number 99960 by bemath last updated on 24/Jun/20
Givenyx+xy=2.findthevalueofdydx∣(1,1)=?
Commented by Dwaipayan Shikari last updated on 24/Jun/20
−1
Answered by Rio Michael last updated on 24/Jun/20
xdydx+y2x+x2ydydx+y=0⇒1dydx∣(1,1)+121+121dydx∣(1,1)+1=0
Answered by 1549442205 last updated on 24/Jun/20
Derivativetwosidesofthegivenweweequationbyxweobtainy′x+y2x+y+xy′2y=0⇒y′(x+x2y)=−(y2x+y)⇒y′=−(2xy+y)2xy+x⇒y′∣(1,1)=−(2xy+y)2xy+x∣(1,1)=−33=−1
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