Question

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

Answer 1

`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`