Q.

If x+siny=logx, then dydx is equal to:

1 1-xxsiny  
2 1-xxcosy  
3 1+xxcosy  
4 None of these  

Ans.

(2)

Ans.      1-xxcosy

Explanation:

        siny+x=logx

On differentiating with respect to x, we get

   cosydydx+1=1x

⇒  cosydydx=1x-1

⇒  cosydydx=1-xx

⇒           dydx=1-xxcosy