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