Using the limit definition, how do you differentiate #f(x)=2x^2-x#?

1 Answer
Johnson Z.
May 14, 2016

#f'(x)=4x-1#

Explanation:

The limit definition of a derivative states that

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

Substituting #f(x)=2x^2-x# into #f'(x)#,

#f'(x)=lim_(hrarr0)((2(x+h)^2-(x+h))-(2x^2-x))/h#

From this point on, you want to expand and simplify.

#f'(x)=lim_(hrarr0)((2x^2+4xh+2h^2-x-h)-2x^2+x)/h#

#f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)(2x^2)+4xh+2h^2color(blue)cancelcolor(black)(-x)-hcolor(red)cancelcolor(black)(-2x^2)color(blue)cancelcolor(black)(+x))/h#

#f'(x)=lim_(hrarr0)(4xh+2h^2-h)/h#

Factor out #h# from the numerator.

#f'(x)=lim_(hrarr0)(h(4x+2h-1))/h#

#f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)h(4x+2h-1))/color(red)cancelcolor(black)h#

#f'(x)=lim_(hrarr0)4x+2h-1#

Plugging in #h=0#,

#f'(x)=lim_(hrarr0)4x+2(0)-1#

#f'(x)=lim_(hrarr0)4x-1#

#color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=4x-1)color(white)(a/a)|)))#

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