What is the equation of the tangent line to the curve y = 2xsin(x) at point (pi/2, pi)?

1 Answer
Oct 10, 2017

#y = 2(x - pi/2) + pi#

Explanation:

STEP 1:
Find the derivative of y (the slope of the tangent line). We need to use the product rule to take this derivative.
Treat 2x as the "first term" and sin(x) as the "second term":

#y' = d/dx(2x) sin(x) + 2xd/dx(sin(x))#

STEP 1a: find the derivative of 2x and sin(x), to plug into the equation

#d/dx(2x) = 2#, and #d/dx(sin(x)) = cos(x)#

Step 1b: plug into equation

#y' = 2sin(x) + 2xcos(x)#

STEP 2: find the derivative at #pi/2#

Since we are looking to find the equation of the tangent line at #x=pi/2#, we need to find the slope at that point:

#y'(pi/2) = 2(sin(pi/2)) + 2(pi/2)(cos(pi/2))#
#=> y'(pi/2) = 2(1) + 2(pi/2)(0)#
#=> y'(pi/2) = 2 + 0#
# => y'(pi/2) = 0 #

STEP 3: plug in the tangent slope found in step 2 and point given into the point-slope equation #[y - y1 = m(x - x1)]#

#y - pi = 2(x - pi/2) => y = 2(x - pi/2) + pi#