I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives
sec²(x) dx/dt = a exp(at)
Recall that
sec²(x) = 1 + tan²(x)
Then we have
(1 + tan²(x)) dx/dt = a exp(at)
(1 + exp(2at)) dx/dt = a exp(at)
dx/dt = a exp(at) / (1 + exp(2at))
If exp(y) = 1 + exp(2at), then differentiating with respect to t yields
exp(y) dy/dt = 2a exp(2at)
(1 + exp(2at)) dy/dt = 2a exp(2at)
dy/dt = 2a exp(2at) / (1 + exp(2at))
By the chain rule,
dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)
Then the first derivative is
dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))
dy/dx = (2a exp(2at)) / (a exp(at))
dy/dx = 2 exp(at)
Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write
dy/dx = f(t)
By the chain rule, the derivative is
d²y/dx² = df/dx
d²y/dx² = df/dt • dt/dx
d²y/dx² = (df/dt) / (dx/dt)
d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))
d²y/dx² = 2 (1 + exp(2at))
