How do you find the derivative of #ln((sin^2)x) #? Calculus Differentiating Trigonometric Functions Intuitive Approach to the derivative of y=sin(x) 1 Answer lvoyster May 9, 2018 #d/(dx) ln(sin^2(x)) =2 cot(x)# Explanation: #d/(dx) ln(sin^2(x)) = (d/(dx)sin^2(x))/(sin^2(x))=(2 sin(x) cos(x))/(sin^2(x))=(2 cos(x))/(sin(x))# 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 2013 views around the world You can reuse this answer Creative Commons License