Using the limit definition, how do you find the derivative of # f(x) = 2x^2-x#?
1 Answer
Explanation:
The limit definition of the derivative states that the derivative of the function
#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#
Here,
The function's derivative is
#f'(x)=lim_(hrarr0)([2(x+h)^2-(x+h)]-(2x^2-x))/h#
Distribute and simplify.
#=lim_(hrarr0)([2(x^2+2xh+h^2)-x-h]-2x^2+x)/h#
#=lim_(hrarr0)([2x^2+4xh+2h^2-x-h]-2x^2+x)/h#
#=lim_(hrarr0)(color(red)(cancel(color(black)(2x^2)))+4xh+2h^2color(red)(cancel(color(black)(-x)))-hcolor(red)(cancel(color(black)(-2x^2)))color(red)(cancel(color(black)(+x))))/h#
#=lim_(hrarr0)(4xh+2h^2-h)/h#
Divide an
#=lim_(hrarr0)4x+2h-1#
Now the limit can be evaluated by plugging in
#f'(x)=4xcolor(red)(cancel(color(black)(+2(0))))-1#
#f'(x)=4x-1#