For #f(x)=xsin^3(x/3)# what is the equation of the tangent line at #x=pi#?

Question

Ezra Adams · Accepted Answer

#y=1.8276x-3.7#

You have to find the derivative:

#f'(x)=(x)'sin^3(x/3)+x*(sin^3(x/3))'#

In this case, the derivative of the trigonometric function is actually a combination of 3 elementary functions. These are:

#sinx#

#x^n#

#c*x#

The way this will be solved is as follows:

#(sin^3(x/3))'=3sin^2(x/3)*(sin(x/3))'=#

#=3sin^2(x/3)*cos(x/3)(x/3)'=#

#=3sin^2(x/3)*cos(x/3)*1/3=#

#=sin^2(x/3)*cos(x/3)#

Therefore:

#f'(x)=1*sin^3(x/3)+x*sin^2(x/3)*cos(x/3)#

#f'(x)=sin^3(x/3)+x*sin^2(x/3)*cos(x/3)#

#f'(x)=sin^2(x/3)*(sin(x/3)+xcos(x/3))#

The derivation of the tangent equation:

#f'(x_0)=(y-f(x_0))/(x-x_0)#

#f'(x_0)*(x-x_0)=y-f(x_0)#

#y=f'(x_0)*x-f'(x_0)*x_0+f(x_0)#

Substituting the following values:

#x_0=π#

#f(x_0)=f(π)=π*sin^3(π/3)=2.0405#

#f'(x_0)=f'(π)=sin^2(π/3)*(sin(π/3)+πcos(π/3))=1.8276#

Therefore, the equation becomes:

#y=1.8276x-1.8276*π+2.0405#

#y=1.8276x-3.7#

In the graph below you can see that at #x=π=3.14# the tangent is indeed increasing and will intersect the y'y axis at #y<0#

graph{x(sin(x/3))^3 [-1.53, 9.57, -0.373, 5.176]}

Charles Arocha · Answer

undefined The equation of the tangent line at x=pi for the function f(x)=xsin^3(x/3) is y = -pi/3 + 3pi/2.