What is the equation of the tangent line of #f(x)=Arctan(x/3) # at #x=0#?

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Answer 1 · 2016-09-08T09:51:44

Equation of desired tangent is #3y=x#

At #x=0#, #f(x)=arctan(x/3)# has a value of #arctan0=0#, hece we are seeking a tangent at #(0,0)#.

The slope of the tangent will be given by #f'(0)#.

As #d/(dx)arctanx=1/(1+x^2)#

#f'(x)=1/(1+(x/3)^2)xx1/3=9/(9+x^2)xx1/3=3/(9+x^2)#

and at #x=0#, #f'(0)=3/9=1/3#

Hence the tangent has a slope of #1/3# and passes through #(0.0)#.

As equation of a line with slope #m# and passing trough #(x_1,y_1)# is given by

#(y-y_1)=m(x-x_1)#

Equation of desired tangent is #y-0=1/3(x-0)# or #3y=x#

graph{(y-arctan(x/3))(3y-x)=0 [-5, 5, -2.5, 2.5]}