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

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Answer 1 · 2016-01-05T03:43:14

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

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.