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

Question

Emilia Aguilar · Accepted Answer

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

The limit definition of the derivative states that the derivative of the function #f(x)# is

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

Here, #f(x)=2x^2-x#, so #f(x+h)=2(x+h)^2-(x+h)#.

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 #h# from each term.

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

Now the limit can be evaluated by plugging in #0# for #h#.

#f'(x)=4xcolor(red)(cancel(color(black)(+2(0))))-1#

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

Emilia Aguilar · Answer

undefined To find the derivative of $ f(x) = 2x^2 - x $ using the limit definition of the derivative, we use the formula:

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

First, we substitute the function $ f(x) = 2x^2 - x $ into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{2(x+h)^2 - (x+h) - (2x^2 - x)}{h} $$

Next, we expand and simplify the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) - x - h - 2x^2 + x}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 - x - h - 2x^2 + x}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 - h}{h} $$

$$ f'(x) = \lim_{h \to 0} 4x + 2h - 1 $$

Finally, we evaluate the limit as $ h $ approaches 0:

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

So, the derivative of $ f(x) = 2x^2 - x $ is $ f'(x) = 4x - 1 $.