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)
= 2pisqrt(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 = [2pisqrt(2)+4(sqrt(2))](pi/2) + c
c = -1.98
Thus, the equation is:
y = 14.5x - 1.98
