How do you differentiate #f(x)=1/x^3-x^4+sinx# using the sum rule? Calculus Basic Differentiation Rules Sum Rule 1 Answer Anees Apr 7, 2016 #f'(x)=-3/(x^4)-4x^3+cosx# Explanation: #f(x)=1/(x^3)-x^4+sinx# #=>f(x)=x^-3-x^4+sinx# Differentiating both sides w.r.t 'x' #f'(x)=d/(dx)(x^-3-x^4+sinx)# #f'(x)=d/(dx)(x^-3)-d/(dx)(x^4)+d/(dx)(sinx)# #f'(x)=-3x^-4-4x^3+cosx# #f'(x)=-3/(x^4)-4x^3+cosx# Answer link Related questions What is the Sum Rule for derivatives? How do you find the derivative of #y=f(x)+g(x)#? How do you find the derivative of #y = f(x) - g(x)#? What is the derivative of #f(x) = xlnx-lnx^x#? How do you differentiate #f(x)=1/x+1/x^3# using the sum rule? How do you differentiate #f(x)=x+x-2x# using the sum rule? How do you differentiate #f(x)=x^2-x-x(x-1)# using the sum rule? How do you differentiate #f(x)=x^3-x^2+4x-1# using the sum rule? How do you differentiate #f(x)=sinx+cosx-x^3# using the sum rule? How do you differentiate #f(x)=x+lnx^2-x^2# using the sum rule? See all questions in Sum Rule Impact of this question 2182 views around the world You can reuse this answer Creative Commons License