How do you use the definition of a derivative to find the derivative of #f(x) = -7x^2 + 4x#?

Question

Cameron Alexander · Accepted Answer

#f'(x)=-14x+4# By definition: #f'(x) = lim_"h->0" (f(x+h)-f(x))/h# In this example: #f(x) = -7x^2+4x# Therefore: #f'(x) = lim_"h->0" (-7(x+h)^2+4(x+h) -(-7x^2+4x))/h# #= lim_"h->0" (-7(x^2+2xh+h^2)+4x+4h+7x^2-4x)/h# #= lim_"h->0" (-7x^2-14xh-7h^2+4x+4h+7x^2-4x)/h# #= lim_"h->0" (cancel(-7x^2)-14xh-7h^2cancel(+4x)+4hcancel(+7x^2)cancel(-4x))/h# #=lim_"h->0" (-14xh-7h^2+4h)/h# #=lim_"h->0" -14x-7h+4# #= -14x+4#

Charles Anctil · Answer

undefined To find the derivative of $ f(x) = -7x^2 + 4x $ using the definition of a derivative, follow these steps:

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

2. Substitute $ f(x) = -7x^2 + 4x $ into the definition:
   $ f'(x) = \lim_{h \to 0} \frac{[-7(x + h)^2 + 4(x + h)] - (-7x^2 + 4x)}{h} $

3. Expand and simplify the expression inside the limit:
   $ f'(x) = \lim_{h \to 0} \frac{[-7(x^2 + 2hx + h^2) + 4x + 4h] + 7x^2 - 4x}{h} $
   $ f'(x) = \lim_{h \to 0} \frac{-7x^2 - 14hx - 7h^2 + 4x + 4h + 7x^2 - 4x}{h} $
   $ f'(x) = \lim_{h \to 0} \frac{-14hx - 7h^2 + 4h}{h} $

4. Cancel out common terms and factor out $ h $:
   $ f'(x) = \lim_{h \to 0} \frac{h(-14x - 7h + 4)}{h} $

5. Simplify further:
   $ f'(x) = \lim_{h \to 0} -14x - 7h + 4 $

6. Evaluate the limit as $ h $ approaches 0:
   $ f'(x) = -14x + 4 $

Therefore, the derivative of $ f(x) = -7x^2 + 4x $ is $ f'(x) = -14x + 4 $.