What is the equation of the line tangent to #f(x)=x sin2x # at #x=pi/12#?

1 Answer
Sep 27, 2017

#y - pi/24 = ((6+pi*sqrt(3))/12)(x - pi/12)#

Explanation:

To determine the equation of the tangent line, we will require a slope and a point, given that #x = pi/12#. The slope will come from the derivative #f'(x)#, which is found with the Product Rule and Chain Rule:

#f'(x) = (x) * ( cos(2x) * 2) + (sin(2x)) * (1)#
#f'(x) = 2x cos(2x) + sin(2x)#

At #x = pi/12#, #f'(pi/12) = 2(pi/12)*cos(2*pi/12) + sin(2*pi/12)#
# = pi/6 * cos(pi/6) + sin(pi/6) = pi/6 * sqrt(3)/2 + 1/2#
# = (pi*sqrt(3))/12 + 1/2 = (pi*sqrt(3))/12 + 6/12 = (6+pi*sqrt(3))/12#

To find the point of tangency, we use #f(x)# to find the #y# value when #x = pi/12#:

#f(pi/12) = (pi/12) * sin(2*pi/12) = pi/12 * sin(pi/6) = pi/12 * 1/2 = pi/24#

We can then use the point-slope formula to determine the equation of the tangent line:

#y - y_1 = m(x - x_1)#
#y - pi/24 = ((6+pi*sqrt(3))/12)(x - pi/12)#