Using the limit definition, how do you find the derivative of # f(x) = 4 -2x -x^2#? Calculus Derivatives Limit Definition of Derivative 1 Answer Andrea S. May 13, 2018 #d/dx (4-2x-x^2) = -2-2x# Explanation: By definition: #d/dx (4-2x-x^2) = lim_(h->0) ( ( 4-2(x+h) -(x+h)^2 ) - (4-2x-x^2))/h# #d/dx (4-2x-x^2) = lim_(h->0) ( ( color(blue)(4) color(red)(-2x)-2h color(green)(-x^2)-2hx -h^2 color(blue)(- 4) color(red)(+2x) color(green)(+x^2)))/h# #d/dx (4-2x-x^2) = lim_(h->0) ( -2h-2hx-h^2)/h# #d/dx (4-2x-x^2) = lim_(h->0) ( -2-2x-h)# #d/dx (4-2x-x^2) = -2-2x# Answer link Related questions What is the limit definition of the derivative of the function #y=f(x)# ? Ho do I use the limit definition of derivative to find #f'(x)# for #f(x)=3x^2+x# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=sqrt(x+3)# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=1/(1-x)# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=x^3-2# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=1/sqrt(x)# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=5x-9x^2# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=sqrt(2+6x)# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=mx+b# ? How do I use the limit definition of derivative to find #f'(x)# for #f(x)=c# ? See all questions in Limit Definition of Derivative Impact of this question 1935 views around the world You can reuse this answer Creative Commons License