How do you find the derivative of #f(x)=2x^2+x-1# using the limit process?

Question

Charles Anctil · Accepted Answer

Based on the limit definition of derivative: #(df)/dx = lim_(h->0) (f(x+h)-f(x))/h# #(df)/dx = lim_(h->0) ( (2(x+h)^2+x+h-1) - (2x^2+x-1))/h# #(df)/dx = lim_(h->0) ( cancel(2x^2)+4xh+2h^2+cancel(x)+h -cancel(1) - cancel(2x^2)-cancel(x)+cancel(1))/h# #(df)/dx = lim_(h->0) ( 4xh+2h^2+h )/h# #(df)/dx = lim_(h->0) 2h+4x+1# #(df)/dx = 4x+1#

Paisley Johnson · Answer

undefined To find the derivative of $f(x) = 2x^2 + x - 1$ using the limit process, you can 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 the given function $f(x) = 2x^2 + x - 1$ into the definition:
   $$f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 + (x+h) - 1) - (2x^2 + x - 1)}{h}$$

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

4. Factor out $h$ from the numerator and cancel it with $h$ in the denominator:
   $$f'(x) = \lim_{h \to 0} \frac{h(4x + 2h + 1)}{h}$$

5. Simplify the expression:
   $$f'(x) = \lim_{h \to 0} (4x + 2h + 1)$$

6. Now, take the limit as $h$ approaches 0:
   $$f'(x) = 4x + 1$$

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