How do you find the fist and second derivative of #pi*sin(pix)#? Calculus Differentiating Trigonometric Functions Intuitive Approach to the derivative of y=sin(x) 1 Answer Massimiliano Apr 30, 2015 In this way: #y'=pi*(cospix*pi)=pi^2cospix# #y''=pi^2(-sinpix*pi)=-pi^3sinpix#. 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)#? If f(x)= 2x sin(x) cos(x), how do you find f'(x)? What is the derivative of #sin^2 (6x)#? See all questions in Intuitive Approach to the derivative of y=sin(x) Impact of this question 5955 views around the world You can reuse this answer Creative Commons License