Question

If x = sin^–1(e^t), y = `sqrt(1 - e^(2t)), "show that"  sin x + dy/dx` = 0

Answer 1

x = sin^–1(e^t), y = `sqrt(1 - e^(2t))`
Differentiating x and y w.r.t. t, we get
`dx/dt = d/dt[sin^-1(e^t)]`

= `(1)/sqrt(1 - (e^t)^2).d/dt(e^t)`

= `(1)/sqrt(1 - e^(2t)) xx e^t = e^t/sqrt(1 - e^(2t))`  and
`dy/dt = d/dt(sqrt(1 - e^(2t)))`

= `(1)/(2sqrt(1 - e^(2t))).d/dt(1 - e^(2t))`

= `(1)/(2sqrt(1 - e^(2t))). xx (0 - e^(2t) xx 2)`

= `(-e^(2t))/sqrt(1 - e^(2t))`

∴ `dy/dx = ((dy/dt))/((dx/dt)`

= `(((-e^(2t))/sqrt(1 - e^(2t))))/(((e^t)/sqrt(1 - e^(2t)))`
= – e^t
= – sin x                ...[∵ x = sin^–1(e^t)]
∴ `sin x + dy/dx` = 0.