Find `(dy)/(dx),w h e n` `x=e^(theta)(theta+1/theta)a n dy=e^(-theta)(theta-1/theta)`

Answer

N/a

Explanation

`:. x = e^(theta)(theta+(1)/(theta))` and `y=e^(-theta)(theta-(1)/(theta))`
`:. (dx)/(d theta) = (d)/(d theta)[e^(theta).(theta+(1)/(theta))]`
`=e^(theta).(d)/(d theta)(theta+(1)/(theta))+(theta+(1)/(theta)).(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)/(d theta) = (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)(1+(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)`
`:. (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^(-2theta)((-theta^(3)+theta^(2)+theta+1)/(theta^(3)+theta^(2)+theta-1))`