Find the equation of the tangent line, in standard form, to the curve y = sin4xcos^2 x when x = 0.

Answer 1

y = sin4xcos^2 x ???
Assuming you mean
y = sin(4x)(cos^2 x)
y = sin(4x) (cosx)^2
dy/dx = sin(4x) (2cosx)(-sinx) + (cosx)^2 (4cos (4x))

when x = 0 , y = 0, so the tangent passes through (0,0)
when x = 0, dy/dx = 0 + 1(4) = 4

tangent equation at (0,0) is
y = 4x