What is the derivative of #sin(x(pi/8))#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Anthony R. Jan 10, 2018 #y'=d/dx[sin((pix)/8)]=pi/8cos((pix)/8)# Explanation: Apply the Chain Rule: #y'=d/dx[f(g(x)]=f'(g(x))*g'(x)# Given: #sin(x(pi/8))#, We can rewrite this as #sin((pix)/8)# Let #f(x)=sinx# and #g(x)=(pix)/8# Thus, #f'(x)=cosx# and #g'(x)=pi/8# So, #y'=d/dx[sin((pix)/8)]=cos((pix)/8)*pi/8# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 2906 views around the world You can reuse this answer Creative Commons License