Find `(dy)/(dx)` , when `x=e^(theta)(theta+1/theta)` and `y=e^(-theta)(theta-1/theta)`

Options

A. `e^(2 theta) ((- theta ^(3) + theta ^(2) + theta + 1)/( theta ^(3) + theta ^(2) + theta - 1)) `
B. `e^(- 2 theta) ((- theta ^(3) + theta ^(2) + theta + 1)/( theta ^(3) + theta ^(2) + theta - 1)) `
C. `e^(-2 theta) ((- theta ^(3) + theta ^(2) + theta + 1)/( -theta ^(3) + theta ^(2) + theta - 1)) `
D. None of the above

Answer

A

Explanation

Given , `x = e^(theta) ( theta + (1)/(theta))" and " y = e^(-theta)( theta - (1)/(theta))`
` (dx)/(xtheta) = (d)/(d theta) [e^(theta).( theta - (1)/(theta))]`
`= e^(theta) (d)/(d theta) (theta- (1)/(theta)) + (theta - (1)/(e)) . (d)/(d theta )e^(theta)`
` e^(theta) (1 - (1)/(theta ^(2)))+ ( theta - (1)/(theta)) . e^(theta) `
` = e^(theta) (1 - (1)/(theta^(2)) + theta - (1)/(theta)) = e^(theta) ((theta^(2) - 1 + theta ^(3) + theta)/(theta ^(2))) `...(i)
and `(dy)/(dx) = (d)/(d theta) [ e^(-theta)(theta - (1)/(theta))]`
` = e^(-theta) (d)/(d theta)(theta- (1)/(theta) )+(d)/(d theta) e^(theta)( theta - (1)/(theta))`
` = e^(-theta) (theta- (1)/(theta^(2)) )+( theta - (1)/(theta)) e^(-theta). (d)/(d theta ) (- theta )`
` = e^(-theta) [(theta^(2) +1)/(theta ^(2) ) - (theta^(2)-1)/(theta)] = e^(-theta) [(theta^(2) +1- theta^(3) + theta )/(theta ^(2) )]`...(ii)
` therefore (dy)/(dx) = (dy//(d theta ))/(dx//(d theta )) = ( e^(- theta) (( theta ^(2) + 1 - theta ^(3) + theta)/(theta^(2))))/( e^(- theta) (( theta ^(2) - 1 - theta ^(3) + theta)/(theta^(2))))`
` e^(- theta) (( -theta ^(3) + theta ^(2) + theta+1)/(theta ^(3) + theta^(2)+ theta - 1))`