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 2348 views around the world You can reuse this answer Creative Commons License