Given: Find tangent line of #f(x) = (tan^3x)/x " at " x = pi/4#
slope #= m = f'(pi/4)#
Find the first derivative of #f(x)#:
Use the Quotient Rule: #f'(x) = (u/v)' = (v u' - u v')/v^2#
From #f(x) = (tan^3x)/x; " Let " u = tan^3x = (tan x)^3; " " v = x#
Use the Power Rule: #(w^n)' = n(w)^(n-1) w'#
#w = tan x; " "w' = sec^2x; " " n = 3#
#u' = (w^3)' =3 (tan x)^2 sec^2 x; " " v' = 1#
#f'(x) = (3xtan^2x sec^2x - tan^3x)/x^2 = (3tan^2x sec^2x)/x - (tan^3x)/x^2#
Find the slope at #pi/4 (45^o)#:
#tan (pi/4) = 1; " " sec (pi/4) = 1/(cos(pi/4)) = 1/(1/sqrt(2)) = sqrt(2)#
#sec (pi/4)^2 = 2#
#m = f'(pi/4) = (3*1*2)/(pi/4) - 1/(pi^2/16) = 24/pi - 16/(pi^2) = (24 pi -16)/pi^2 #
#m = (8(3pi-2))/pi^2#
Find location of the tangent point:
#f(pi/4) = (tan^3(pi/4))/(pi/4) = 1^3/(pi/4) = 4/(pi); " " (pi/4, 4/(pi))#
You can find the tangent line different ways, using the slope intercept form of the line or the point-slope form.
Using #y = mx + b# plug the tangent point into the equation and solve for #b#, the #y-#intercept:
#4/(pi) = (24/pi - 16/(pi^2))pi/4 + b#
#4/(pi) = (6 - 4/(pi)) + b#
#b = 4/(pi) - (6 - 4/(pi)) = 4/(pi) -6 + 4/(pi) = 8/(pi) - 6 = (8-6pi)/pi#
#b = (2(4-3pi))/pi#
Tangent line: #y = (8(3pi -2))/pi^2 x + (2(4-3pi))/pi#
