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