If #y=sin^5x# then what is #dy/dx#? Calculus Differentiating Trigonometric Functions Intuitive Approach to the derivative of y=sin(x) 1 Answer Steve M Oct 4, 2016 Use the chain rule Explanation: If #y=sin^5x#, then Let #u=sin x#, Then #(du)/(dx)=cosx# and #y=u^5# so #(dy)/(du)=5u^4# Chain rule gives #(dy)/(dx)=(dy)/(du) (du)/(dx)#. so: #(dy)/(dx)= 5u^4 cosx = 5sin^4x cos 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 13551 views around the world You can reuse this answer Creative Commons License