What is the first derivative of #x^2 + (16/x)#? Calculus Basic Differentiation Rules Power Rule 1 Answer James May 13, 2018 #y'=2x-16/x^2# Explanation: show below #y=x^2 + (16/x)# #y'=2x-16/x^2# 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 4870 views around the world You can reuse this answer Creative Commons License