How do you differentiate #cosx/(sinx-2)#? Calculus Differentiating Trigonometric Functions Intuitive Approach to the derivative of y=sin(x) 1 Answer Anjali G Nov 11, 2016 #f(x)=(cosx)/(sinx-2)# Use quotient rule: #f'(x)=[(sinx-2)(-sinx)-(cosx)(cosx)]/[(sinx-2)^2]# #f'(x)=(-sin^2x+2sinx-cos^2x)/(sinx-2)^2# #f'(x)=(2sinx-(sin^2x+cos^2x))/(sinx-2)^2# #f'(x)=(2sinx-1)/(sinx-2)^2# Answer link Related questions What is the derivative of #-sin(x)#? What is the derivative of #sin(2x)#? How do I find the derivative of #y=sin(2x) - 2sin(x)#? How do you find the second derivative of #y=2sin3x-5sin6x#? How do you compute #d/dx 3sinh(3/x)#? How do you find the derivative #y=xsinx + cosx#? What is the derivative of #sin(x^2y^2)#? What is #f'(-pi/3)# when you are given #f(x)=sin^7(x)#? How do you find the fist and second derivative of #pi*sin(pix)#? If f(x)= 2x sin(x) cos(x), how do you find f'(x)? See all questions in Intuitive Approach to the derivative of y=sin(x) Impact of this question 1823 views around the world You can reuse this answer Creative Commons License