A) \[\frac{x}{y}\]
B) \[\frac{y}{x}\]
C) \[y\]
D) \[x\]
E) \[\frac{x}{a}\]
Correct Answer: B
Solution :
\[sec\left( \frac{x+y}{x-y} \right)=a\] \[\Rightarrow \] \[\left( \frac{x+y}{x-y} \right)={{\sec }^{-1}}(a)\] On differentiating with respect to\[x\] \[\Rightarrow \] \[\frac{(x-y)\left( 1+\frac{dy}{dx} \right)-(x+y)\left( 1-\frac{dy}{dx} \right)}{{{(x-y)}^{2}}}=0\] \[\Rightarrow \] \[x+x\frac{dy}{dx}-y-y\frac{dy}{dx}-x+x\frac{dy}{dx}-y\] \[+y\frac{dy}{dx}=0\] \[\Rightarrow \] \[2x\frac{dy}{dx}=2y\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{x}\]
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