What is the equation of the line tangent to f(x)=x ^2cos2x at x=pi/4?

1 Answer
Feb 28, 2016

y=(-pi^2/8)x+pi^3/32

Explanation:

We will first find the derivative of f(x).

[1]" "f'(x)=d/dx(x^2cos2x)

By product rule, this will become:

[2]" "f'(x)=d/dx(x^2)*cos2x+d/dx(cos2x)*x^2

By power rule, d/dx(x^2) becomes 2x^1.

[3]" "f'(x)=2x*cos2x+d/dx(cos2x)*x^2

The derivative of cosx is -sinx. But since we are getting the derivative of cos2x instead, we need to use chain rule.

[4]" "f'(x)=2x*cos2x+(-sin2x)*d/dx(2x)*x^2

The derivative of 2x is simply 2.

[5]" "f'(x)=2x*cos2x+(-sin2x)*2*x^2

[6]" "f'(x)=2xcos2x-2x^2sin2x


Now that we have the derivative of f(x), we can solve for the slope of the tangent line at x=pi/4.

[1]" "m=2(pi/4)cos(2*pi/4)-2(pi/4)^2sin(2*pi/4)

[2]" "m=(pi/2)cos(pi/2)-(pi^2/8)sin(pi/2)

[3]" "m=(pi/2)(0)-(pi^2/8)(1)

[4]" "m=-pi^2/8


We now look for the point of f(x) at x=pi/4.

[1]" "f(pi/4)=(pi/4)^2cos(pi/2)

[2]" "f(pi/4)=(pi^2/16)(0)

[3]" "f(pi/4)=0

[4]" "(x_0,y_0)=(pi/4,0)


Finally, we can solve for the equation of the tangent line.

[1]" "y-y_0=m(x-x_0)

[2]" "y-0=(-pi^2/8)(x-pi/4)

[3]" "color(red)(y=(-pi^2/8)x+pi^3/32)