A) \[\frac{1}{2y+1}\]
B) \[\frac{1}{2y-1}\]
C) \[\frac{1}{xy}\]
D) 1
Correct Answer: B
Solution :
Let \[y=\sqrt{x+\sqrt{x+\sqrt{x+....\infty }}}\] \[\Rightarrow \] \[y=\sqrt{x+y}\] \[\Rightarrow \] \[{{y}^{2}}=x+y\] \[\Rightarrow \] \[{{y}^{2}}-y-x=0\] On differentiating w.r.t. \[x,\]we get \[2y\frac{dy}{dx}-\frac{dy}{dx}-1=0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{(2y-1)}\]You need to login to perform this action.
You will be redirected in
3 sec