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

Question

Paisley Johnson · Accepted Answer

#f'(x)=4x-1# By limit definition, the derivative of any function is #f'(x)=lim_(h o0) (f(x+h)-f(x))/h# Now, for this particular question, we have #f(x)=2x^2-x# So, using the derivative definition, we have #f'(x)=lim_(h o0)((2(x+h)^2-(x+h))-(2x^2-x))/h# Rearranging the equation and separating to have the #x# parameters of same power, we have #f'(x)=lim_(h o0)((2(x+h)^2-2x^2)/h-(x+h-x)/h)# In the right side, the second term is simple so we'll directly get it as #1#. In the first term, notice that we can take #2# as a common term. Next, by taking #a=x+h# and #b=x#, we can use the identity #a^2-b^2=(a+b)(a-b)# for which we get the answer #(2x+h)(h)# Substituting that above, we get #f'(x)=lim_(h o0)(2(2x+h)h/h-1)=lim_(h o0)(2(2x+h)-1)# Now, since it's not possible to reduce the terms any further, we can finally just take #h=0# and get what we wanted. Hence you should get the answer I've written way above.

Christopher Agresta · Answer

undefined To find the derivative of $ f(x) = 2x^2 - x $ using the limit definition, follow these steps:

1. Start with the definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

2. Substitute $ f(x) = 2x^2 - x $ into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{2(x+h)^2 - (x+h) - (2x^2 - x)}{h} $$

3. Expand and simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) - x - h - 2x^2 + x}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 - x - h - 2x^2 + x}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 - h}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{h(4x + 2h - 1)}{h} $$

4. Cancel out the $ h $ in the numerator and denominator:
$$ f'(x) = \lim_{h \to 0} (4x + 2h - 1) $$

5. 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 $.