What is the derivative of #g(t)=(pi)(cos t - 1/t^2)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer jim p. Aug 31, 2017 #=pi(2/(t^3) - sin(t))# Explanation: remember that #pi# is a constant, so you can just mutiply through: #g(t) = pi*cos(t) - pi/(t^2) = pi * cos(t) - pi * t^-2# #(dg)/dt = -pi sin(t) + 2pit^-3# #=pi(2/(t^3) - sin(t))# 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 1375 views around the world You can reuse this answer Creative Commons License