How do you find the f'(x) using the formal definition of a derivative if #f(x)= 2x^2 - 3x+4#?

Question

How do you find the f'(x) using the formal definition of a derivative if #f(x)=  2x^2 - 3x+4#?

Christopher Allers · Accepted Answer

#f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h# For #f(x) = 2x^2-3x+4#
this becomes #f'(x)# #= lim_(hrarr0)((2(x+h)^2 -3(x+h) +4) - (2x^2 -3x +4))/h# #= lim_(hrarr0) (4xh + h^2-3h)/h# #=lim_(hrarr0) (4x +h -3)# #= 4x-3#

Emily Ammerman · Answer

undefined To find $ f'(x) $ using the formal definition of a derivative if $ f(x) = 2x^2 - 3x + 4 $:

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

2. Substitute the function $ f(x) = 2x^2 - 3x + 4 $ into the definition:
   $ f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 - 3(x+h) + 4) - (2x^2 - 3x + 4)}{h} $.

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

5. Cancel common terms:
   $ f'(x) = \lim_{h \to 0} 4x + 2h - 3 $.

6. Evaluate the limit as $ h $ approaches 0:
   $ f'(x) = 4x - 3 $.

Therefore, the derivative of $ f(x) = 2x^2 - 3x + 4 $ with respect to $ x $ is $ f'(x) = 4x - 3 $.