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