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Question

The derivative of ln(x+sinx) with respect to (x+cosx) is

A
1+cosx(x+sinx)(1sinx)
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B
1cosx(x+sinx)(1+sinx)
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C
1cosx(xsinx)(1+cosx)
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D
1+cosx(xsinx)(1cosx)
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Solution

The correct option is A 1+cosx(x+sinx)(1sinx)
Let u=ln(x+sinx) and v=x+cosx
Now,
dudx=1x+sinx(1+cosx) .... (i)
and dvdx=1sinx ..... (ii)
Now, we can find derivative of u with respect to v by dividing (i) by (ii), we get
du/dxdv/dx=(1+cosx)(x+sinx)(1sinx)
or, dudv=1+cosx(x+sinx)(1sinx)
Hence, A is the correct option.

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