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Question

The value of dydx at x=π2, where y is given by y=xsinx+x, is

A
1+12π
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B
1
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C
12π
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D
112π
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Solution

The correct option is B 1+12π
Since, y=xsinx+x
Let y1=xsinx and y2=x
Taking log on both sides, we get
logy1=sinxlogx
On differentiating w.r.t.x, we get
1y1.dy1dx=cosxlogx+1xsinx
dy1dx=xsinx[cosxlogx+1xsinx]
(dy1dx)x=π2=(π2)sinπ2[cosπ2logπ2+2πsinπ2]
=π2×2π=1
Now, y2=x
On differentiating w.r.t.x, we get
dy2dx=12x
(dy2dx)x=π2=12π2=12π
Since, y=y1+y2
dydx=dy1dx+dy2dx
=1+12π

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