A) 1
B) -1
C) 2
D) 0
Correct Answer: A
Solution :
Since to find \[\frac{dy}{dx}\]at \[x=0\] \[\therefore \] At \[x=0\] \[\Rightarrow \] \[\log (y)=0\Rightarrow y=1\] \[\therefore \]To find \[\frac{dy}{dx}\]at \[(0,1).\] On differentiating, \[\log (x+y)=2xy\]on both sides we get \[\frac{1}{x+y}\left( 1+\frac{dy}{dx} \right)=2x\frac{dy}{dx}+2y.1\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2y(x+y)-1}{1-2(x+y)x}\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(0,1)}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
