If 2x+2y=2x+y, then the value of dydx at x=y=1 is.
A
0
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B
−1
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C
1
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D
2
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Solution
The correct option is B−1 Given, 2x+2y=2x+y On differentiating w.r.t. x, we get 2x(log2)+2y(log2)dydx =2x+y(log2)(1+dydx) ⇒2x+2ydydx=2x+y+2x+y(dydx) ⇒dydx(2y−2x+y)=2x+y−2x ⇒dydx=2x+y−2x2y−2x+y Therefore, (dydx)x=y=1=22−22−22=2−2=−1