find the equation of the line perpendicular to the curve y= (tanx)/(1+tanx) at x= pi/4 ( we can write in terms of pi )

Given that y= tanx/ (1+
tanx)

We need to find the line perpendicular to the tangent
line of y at x= pi/4

Then we will find the tangent
point.

==> x= pi/4 ==> y= tanpi/4 / (1+
tanpi/4) =  1/ (1+1) = 1/2

Then the tangent point is (
pi/4, 1/2)

Now we need to find the
slope.

We will find the slope of the tangent line
first.

Let us differentiate
y.

==> y' = ( tanx)'(tanx+1) - (tanx+1)'*tanx /
(tanx+1)^2

==> y' = ( sec^2 x (tanx+1) - sec^2
x*tanx / (tanx+1)^2

Now we will subsitute with x= pi/4 to
find the slope.

==> y'(pi/4) = [2 (1+1) - 2*1]/
(1+1)^2

==> y' (pi/4) = ( 4 -2)/4 = 2/4 =
1/2

Then the slope of the tangent line if
1/2

Then the slope of the perpendicular line is
-2.

Now we will find the
equation.

==> y-y1 =
m(x-x1)

==> y- 1/2 = -2 (
x-pi/4)

==> y= -2x + pi/2 +
1/2

==> y= -2x +
(pi+1)/2