A) \[\frac{{{y}^{2}}}{{{x}^{2}}}\]
B) \[\frac{y}{x}\]
C) \[\frac{x}{y}\]
D) \[\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]
E) \[0\]
Correct Answer: B
Solution :
\[\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}={{\sec }^{-1}}({{e}^{a}})\] On differentiating w.r.t.\[x,\]we get \[\frac{\left[ \begin{align} & ({{x}^{2}}+{{y}^{2}})\left( 2x-2y\frac{dy}{dx} \right)-({{x}^{2}}-{{y}^{2}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2x+2y\frac{dy}{dx} \right) \\ \end{align} \right]}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}=0\] \[=x({{x}^{2}}+{{y}^{2}})-y({{x}^{2}}+{{y}^{2}})\frac{dy}{dx}\] \[=({{x}^{2}}-{{y}^{2}})x+y({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}\] \[=({{x}^{2}}y-{{y}^{3}}+{{x}^{2}}y+{{y}^{3}})\frac{dy}{dx}\] \[=({{x}^{3}}+x{{y}^{2}}-{{x}^{3}}+x{{y}^{2}})\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2x{{y}^{2}}}{2{{x}^{2}}y}=\frac{y}{x}\]
You need to login to perform this action.
You will be redirected in
3 sec
