Find an equation of the tangent line to the curve
y = 8x sin x at the point (π/2, 4π).
y'=8(sinx)+8xcosx
8(sinπ)+8π(cosπ)
0+(-8π)

Hello
We know that upon differentiating a function at a certain point, we can obtain its gradient at that point. That will also be the gradient of its tangent line. Thus,
y = 8x * sin(x)
y' = 8x * cos(x) + 8sin(x)
Evaluating the gradient at the given point, we insert pi/2 for x.
gradient = 8(pi/2) * cos(pi/2) + 8sin(pi/2)
= 2pi*sqrt(2) + 4(sqrt(2))
The equation of the line will be in the form y = mx + c; where m is the gradient and c is the y intercept. To calculate the y-int, we insert the given x and y values:
4pi = [2pi*sqrt(2)+4(sqrt(2))]*(pi/2) + c
c = -1.98
Thus, the equation is:
y = 14.5x - 1.98

The answer is a. point