Find an equation of the tangent line to the curve y=8x sinx at the point pi/2 4pi

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

Answer: d. 39 cmin the diagram, the approximate length of the minor arc de is 39 centimetres.step-by-step explanation: the formula for the length of an arc in a circle is as displayed below.let the length of the minor arc de = xx = 2( pi )r × angle of sector / 360°from this, we will substitute the values given from the problem into the formula to obtain the length of the minor arc de.x = 2( pi )r × angle of sector / 360°r = radius = 25cmpi as an approximate value = 3.14angle of sector = 120°x = 2( 3.14 )( 25 ) × 120° / 360°x = 157 × 1 / 4x = 39.25x = 39 ( rounded to the nearest whole number as an approximate value )

Im pretty sure it’s the first one