What is the equation of the line normal to f(x)= xsin^2(2x) f(x)=xsin2(2x) at x=pi/8x=π8?

1 Answer
Jan 5, 2016

y-pi/16=(2+pi)/4(x-pi/8)yπ16=2+π4(xπ8)

Explanation:

Use the product rule to find f'(x):

f'(x)=sin^2(2x)*d/dx(x)+xd/dx(sin^2(2x))

Find each derivative individually.

d/dx(x)=1

The next will require the chain rule.

d/dx(sin^2(2x))=2sin(2x)*d/dx(sin(2x))

=2sin(2x) * cos(2x) * d/dx(2x)=4sin(2x)cos(2x)

Plug these back in.

f'(x)=sin^2(2x)+4xsin(2x)cos(2x)

Now, to find the slope of the tangent line at x=pi/8, find f'(pi/8).

f'(pi/8)=sin^2(pi/4)+pi/2sin(pi/4)cos(pi/4)

=1/2+pi/2(1/2)=(2+pi)/4

To find the point the tangent line will intersect, find f(pi/8).

f(pi/8)=pi/8sin^2(pi/4)=pi/8(1/2)=pi/16

The tangent line will intersect the point (pi/8,pi/16) and have a slope of (2+pi)/4.

Relate this information in a line in point-slope form:

y-pi/16=(2+pi)/4(x-pi/8)

graph{(y-pi/16-(2+pi)/4(x-pi/8))(x(sin(2x))^2-y)=0 [-1.495, 2.35, -0.546, 1.376]}

Graphed are the function and its tangent line.