Question

What is the equation of the line tangent to `f(x)=2xcos3x` at `x=pi/3`?

Answer 1

Tangent line:
`y=-2x`

Explanation:

From the given trigonometric equation

`f(x)=2x*cos 3x`

Let `y=2x*cos 3x`
solve for the point `(x_1, y_1)` of which `x_1=pi/3`

`y_1=2(pi/3)*cos 3(pi/3)=(2pi)/3*cos pi=(2pi)/3*(-1)`

`y_1=-(2pi)/3`

The point is `(pi/3, -(2pi)/3)`

Solve for the slope `m=y'(pi/3)`

`y'=2*[x*d/dx(cos 3x)+(cos 3x)d/dx(x)]`
`y'=2*[x*(-3*sin 3x)+(cos 3x)(1)]`

slope `m=y'(pi/3)=2*[(pi/3)*(-3*sin 3(pi/3))+(cos 3(pi/3))(1)]`

`m=2[-pi*sin pi+cos pi]`
`m=2[0-1]`
`m=-2`

Use Point-Slope form to find the equation of the tangent line

`y-y_1=m(x-x_1)`

`y--(2pi)/3=-2(x-pi/3)`

`y+(2pi)/3=-2x+(2pi)/3`

`y=-2x`

God bless....I hope the explanation is useful.