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Question

If y=logx+logx+logx+, then dydx is

A
x2y1
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B
x2y+1
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C
1x(2y1)
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D
1x(12y)
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Solution

The correct option is C 1x(2y1)
y=logx+logx+logx+
y=logx+y
Squaring both the sides, we get
y2=logx+y
Differentiate w r to x
2ydydx=1x+dydx
2ydydxdydx=1x
dydx(2y1)=1x
dydx=1x(2y1)

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