How do you differentiate #root3(x)+root4(x)#? Calculus Basic Differentiation Rules Power Rule 1 Answer Ratnaker Mehta Aug 8, 2016 #=1/(3x^(2/3))+1/(4x^(3/4))#. Explanation: We know that, #d/dx(x^n)=n*x^(n-1)#, and, #d/dx(u+v)=(du)/dx+(dv)/dx# Therefore, #d/dx(root3x+root4x)=d/dxroot3x+d/dxroot4x# #=d/dxx^(1/3)+d/dxx^(1/4)# #=1/3*x^(1/3-1)+1/4*x^(1/4-1)# #=1/(3x^(2/3))+1/(4x^(3/4))#. Answer link Related questions How do you find the derivative of a polynomial? How do you find the derivative of #y =1/sqrt(x)#? How do you find the derivative of #y =4/sqrt(x)#? How do you find the derivative of #y =sqrt(2x)#? How do you find the derivative of #y =sqrt(3x)#? How do you find the derivative of #y =sqrt(x)#? How do you find the derivative of #y =sqrt(x)# using the definition of derivative? How do you find the derivative of #y =sqrt(3x+1)#? How do you find the derivative of #y =sqrt(9-x)#? How do you find the derivative of #y =sqrt(x-1)#? See all questions in Power Rule Impact of this question 1350 views around the world You can reuse this answer Creative Commons License