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Question

If yxxy=1, then the value of dydx at x=1 is

A
2(1ln2)
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B
2(1+ln2)
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C
2ln2
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D
2+ln2
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Solution

The correct option is A 2(1ln2)
We have,
yxxy=1 (1)
At x=1,
y=1+1=2
Now, from (1)
exlnyeylnx=1
On differentiating w.r.t. x, we get
exlny(xydydx+lny)eylnx(dydxlnx+yx)=0
yx(xydydx+lny)xy(dydxlnx+yx)=0
On putting x=1 and y=2, we get
2(12dydx+ln2)(0+2)=0
dydx=22ln2
=2(1ln2)

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