How do you differentiate y=tan(piex-1)?

Question

The answer I was given is pie(secx)^2(piex-1). I can't seem to make this work....any ideas?

Answer 1

y = tan(πx - 1)
let u = πx - 1
now, dy/dx = (dy/du)(du/dx)
d(tan(u)/du = sec^2(u)
du/dx = π, so
dy/dx = πsec^2(u)
dy/dx = πsec^2(πx - 1)

Answer 2

Given the equation y = tan (x* pi - 1). We are to solve for dy / dx. Then by the chain rule for differentiation, we obtain

dy / dx = sec^2(x * pi - 1) * [d(x * pi - 1) / dx ]
         = sec^2(x * pi - 1) * [d(x*pi) /dx - d(1) / dx ]
         = sec^2(x * pi - 1) * [pi - 0 ]
         = sec^2(x * pi - 1) * pi.

Therefore, the derivative of tan (x* pi - 1) is sec^2(x * pi - 1) * pi.

Answer 3

y = tan(pi*x-1)
dy/dx = sec^2(pi*x-1) * d/dx(pi*x-1) - chain rule
dy/dx = sec^2(pi*x-1) * pi

the answer you were given is either wrong or I'm not reading it correctly. MY TI-89 comes up with the same answer that I got. Perhaps you wrote the answer down wrong?

Answer 4

using the chain rule.

d/dx tan(u) = sec^2u * du/dx

u in this case is pi*x - 1

du/dx = pi