Question

What is the equation of the line tangent to `f(x)=x cos^2(2x)` at `x=pi/12`?

Answer 1

`y-pi/16=(9-pisqrt3)/12(x-pi/12)`

Explanation:

First, find the point the tangent line will intercept.

`f(pi/12)=pi/12cos^2(pi/6)=pi/12(sqrt3/2)^2=pi/16`

The point of tangency is `(pi/12,pi/16)`.

To find the slope of the tangent line, find the value of the derivative at `x=pi/12`.

To find the derivative of the function, use the product rule.

`f'(x)=cos^2(2x)d/dx[x]+xd/dx[cos^2(2x)]`

Find each internal derivative:

`d/dx[x]=1`

The next requires the chain rule:

`d/dx[cos^2(2x)]=2cos(2x)d/dx[cos(2x)]`

`=2cos(2x)(-sin(2x))d/dx[2x]=-4cos(2x)sin(2x)`

Thus,

`f'(x)=cos^2(2x)-4xcos(2x)sin(2x)`

The slope of the tangent line is:

`f'(pi/12)=cos^2(pi/6)-pi/3cos(pi/6)sin(pi/6)`

`=3/4+pi/3(sqrt3/2)(1/2)=3/4-(pisqrt3)/12=(9-pisqrt3)/12`

Thus, the tangent line passes through the point `(pi/12,pi/16)` and has a slope of `(9-pisqrt3)/12`.

This can be related in an equation in point-slope form:

`y-pi/16=(9-pisqrt3)/12(x-pi/12)`

Graphed are the function and its tangent line:

graph{(x(cos(2x))^2-y)(y-pi/16-(9-pisqrt3)/12(x-pi/12))=0 [-1.25, 1.787, -0.55, 0.969]}