How do you find the equation of the tangent line of the function #f(x) = sin(2x)# when given the point (#pi,0#)?

Question

How do you find the equation of the tangent line of the function #f(x) = sin(2x)# when given the point (#pi,0#)?

Isaiah Allen · Accepted Answer

#y = 2x-2pi#

The tangent line of a curve #y=f(x)# at a point #(x_1, f(x_1))# has a slope of #f'(x_0)#. Using that, together with the point-slope form of the equation of a line #y-y_1 = m(x-x_1)#, we can solve for the equation of the tangent line at the given point as

#y = f'(x_1)x-x_1f'(x_1)+f(x_1)#

As we are given #(x_1, f(x_1)) = (pi, 0)#, all that remains is to find #f'(pi)#.

Using that #d/dxsin(x) = cos(x)# along with #d/dx cx = c# and the chain rule, we find

#f'(x) = d/dxsin(2x)#

#=cos(2x)(d/dx2x)#
#=2cos(2x)#

Thus #f'(pi) = 2cos(2pi) = 2#

Plugging this into our equation for the tangent line, we get

#y = 2x-pi(2)+0#

#=> y = 2x-2pi#

Graphed together:

William Abdalla · Answer

undefined To find the equation of the tangent line of the function f(x) = sin(2x) at the point (pi,0), we need to find the derivative of the function and then use the point-slope form of a line.

The derivative of f(x) = sin(2x) can be found using the chain rule. The derivative is f'(x) = 2cos(2x).

Now, we can find the slope of the tangent line at the point (pi,0) by substituting x = pi into the derivative. The slope is f'(pi) = 2cos(2pi) = 2cos(0) = 2.

Using the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Substituting (pi,0) and m = 2 into the point-slope form, we get y - 0 = 2(x - pi).

Simplifying the equation, we have y = 2x - 2pi.

Therefore, the equation of the tangent line of the function f(x) = sin(2x) at the point (pi,0) is y = 2x - 2pi.