Find an equation of the tangent line to the curve y = 8x sin(x) at the point (π/2, 4π).

Solution:

Given: Function y = 8x sin(x) and the point (π/2, 4π).

To get the slope of the tangent line, we need to find the derivative of the given function with respect to x and then evaluate the result at the value (π/2, π)

y = 8x sin(x)

dy/dx = [8x][cos x] + [sin x][8]

(slope of tangent line) = m = dy/dx

= [8π/2][cos π/2] + [sin π/2][8]

The slope of the tangent line is

m = [8π/2][0] + [1][8] = 8

⇒ m = 8

We know that slope-point equation y - y₁ = m(x - x₁)

⇒ y - 4π = 8(x - π/2)

Thus, the equation of the tangent line is y - 4π = 8(x - π/2) or 2y - 8π = 8(2x - π)

Find an equation of the tangent line to the curve y = 8x sin(x) at the point (π/2, 4π).

Summary:

The equation of the tangent line to the curve y = 8x sin(x) at the point (π/2, 4π) is 2y - 8π = 8(2x - π).