How do you find f'(x) using the definition of a derivative for #f(x)= 2x^2-x #?

Question

Scarlett Allison · Accepted Answer

#4x-1#
See the explanation.

#f'(x)=lim_(h->0) (f(x+h)-f(x))/h# #lim_(h->0) (2(x+h)^2-(x+h)-(2x^2-x))/h=# #lim_(h->0) (2x^2+4xh+2h^2-x-h-2x^2+x)/h=# #lim_(h->0) (4xh+2h^2-h)/h=lim_(h->0) (4x+2h-1)=4x-1#

Elijah Abram · Answer

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

1. Recall the definition of the derivative: $ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $.
2. Substitute the given function into the definition: $ 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 + 2hx + h^2) - x - h - 2x^2 + x}{h} $.
4. Combine like terms: $ f'(x) = \lim_{h \to 0} \frac{2x^2 + 4hx + 2h^2 - x - h - 2x^2 + x}{h} $.
5. Further simplify: $ f'(x) = \lim_{h \to 0} \frac{4hx + 2h^2 - h}{h} $.
6. Factor out $ h $ from the numerator: $ f'(x) = \lim_{h \to 0} \frac{h(4x + 2h - 1)}{h} $.
7. Cancel out $ h $ from the numerator and denominator: $ f'(x) = \lim_{h \to 0} (4x + 2h - 1) $.
8. Evaluate the limit as $ h $ approaches 0: $ f'(x) = 4x - 1 $.

Therefore, $ f'(x) = 4x - 1 $ is the derivative of $ f(x) = 2x^2 - x $ using the definition of a derivative.