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
Explanation:
Use the product rule to find
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
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)=pi/8sin^2(pi/4)=pi/8(1/2)=pi/16
The tangent line will intersect the point
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.