What is the equation of the tangent line of #f(x) =sinx/cosx-cotx# at #x=pi/4#?

Question

Ezekiel Alexander · Accepted Answer

The equation is #y = 4x - pi#.

Start by finding the point #(x, y)# where the function meets the tangent. #f(pi/4) = sin(pi/4)/cos(pi/4) - cot(pi/4)# #f(pi/4) = sin(pi/4)/cos(pi/4) - cos(pi/4)/sin(pi/4)# #f(pi/4) = (1/sqrt(2))/(1/sqrt(2)) - (1/sqrt(2))/(1/sqrt(2))# #f(pi/4) = 1/sqrt(2) xx sqrt(2)/1 - 1/sqrt(2) xx sqrt(2)/1# #f(pi/4) = 1 - 1# #f(pi/4) = 0# Now, we distinguish the function. #f(x) = sinx/cosx - cosx/sinx# #f'(x) = (cosx xx cosx - (sinx xx -sinx))/(cosx)^2 - (-sinx xx sinx - (cosx xx cosx))/(sinx)^2# #f'(x) = (cos^2x + sin^2x)/cos^2x - (-cos^2x - sin^2x)/sin^2x# #f'(x) = 1/cos^2x - (-(cos^2x + sin^2x))/sin^2x # #f'(x) = sec^2x + csc^2x# Now, the slope of the tangent can be found by evaluating #x = a# within the derivative. #f'(pi/4) = sec^2(pi/4) + csc^2(pi4)# #f'(pi/4) = 1/cos^2(pi/4) + 1/sin^2(pi/4)# #f'(pi/4) = 1/((1/sqrt(2))^2) + 1/(1/sqrt(2))^2# #f'(pi/4) = 1/(1/2) + 1/(1/2)# #f'(pi/4) = 2 + 2# #f'(pi/4) = 4# Now that we are aware of the slope and point of contact, we can determine the tangent line's equation. #y - y_ 1= m(x- x_1)# #y - 0 = 4(x - pi/4)# #y - 0 = 4x - pi# #y = 4x - pi# I hope this is useful!

Charles Arocha · Answer

undefined The equation of the tangent line of f(x) = sinx/cosx - cotx at x = pi/4 is y = -2x + 2√2.