A) \[\frac{y}{x}\]
B) \[-\frac{y}{x}\]
C) \[\frac{x}{y}\]
D) \[-\frac{x}{y}\]
E) \[\frac{x-y}{x+y}\]
Correct Answer: A
Solution :
\[\because \]\[\sec \left( \frac{x-y}{x+y} \right)=a\] \[\Rightarrow \] \[\frac{x-y}{x+y}={{\sec }^{-1}}a\] On differentiating w.r.t.\[x,\]we get \[\frac{(x+y)\left( 1-\frac{dy}{dx} \right)-(x-y)\left( 1+\frac{dy}{dx} \right)}{{{(x+y)}^{2}}}=0\] \[\Rightarrow \] \[x+y-x+y-\{x+y+x-y\}\frac{dy}{dx}=0\] \[\Rightarrow \] \[2y=2x\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{y}{x}\]
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