Using the limit definition, how do you differentiate #f(x) = x^2 - 1598#?

Question

Aurora Alvarado · Accepted Answer

Expand and simplify. Then evaluate the limit.

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h# For this function, we get #f'(x) = lim_(hrarr0)([(x+h)^2-1598] - [x^2-1598])/h# If we try to evaluate the limit by substitution, we get the indeterminate form #0/0#, so we need a different approach. It may not be clear that expanding the numerator will help, but we need to try something, so let's do it and see what happens. #f'(x) = lim_(hrarr0)([x^2+2xh+h^2-1598] - [x^2-1598])/h# # = lim_(hrarr0)(color(red)(x^2)+2xh+h^2color(blue)(-1598) color(red)(-x^2)color(blue)(+1598))/h# # = lim_(hrarr0)(2xh+h^2)/h# If we try substitution, we still get indeterminate form #0/0#. Note however, that for all #h# other that #0#, we have #(2xh+h^2)/h = (cancel(h)(2x+h))/cancel(h) = 2x+h# Therefore, #f'(x) = lim_(hrarr0)(2xh+h^2)/h = lim_(hrarr0)(2x+h)=2x# In many classes it is permissible to write this without comment as follows: #f'(x) = lim_(hrarr0)([(x+h)^2-1598] - [x^2-1598])/h# #= lim_(hrarr0)([x^2+2xh+h^2-1598] - [x^2-1598])/h# # = lim_(hrarr0)(x^2+2xh+h^2-1598 -x^2+1598)/h# # = lim_(hrarr0)(2xh+h^2)/h# # = lim_(hrarr0)(2x+h)# # = 2x#

Cameron Alexander · Answer

undefined To differentiate the function $ f(x) = x^2 - 1598 $ using the limit definition of the derivative, you use the formula:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute $ f(x) = x^2 - 1598 $ into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - 1598 - (x^2 - 1598)}{h} $$

Expand and simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 1598 - x^2 + 1598}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} $$
$$ f'(x) = \lim_{h \to 0} (2x + h) $$

Now, as $ h $ approaches 0, $ 2x + h $ becomes $ 2x $, so:

$$ f'(x) = 2x $$

Thus, the derivative of $ f(x) = x^2 - 1598 $ is $ f'(x) = 2x $.