How do you use the limit definition of the derivative to find the derivative of #f(x)=-4x^2-5x-2#?

Question

Luke Alvarez · Accepted Answer

See below.

We start by finding the gradient in much the same way as we do for any function. namely: #(y_2-y_1)/(x_2-x_1)# If we have a point #P# on a curve with coordinates #(x, f(x))#, then another point #Q# near #P# has coordinates #(x+deltax, f(x+delta x))#, where #delta x # is a small increment of #x#. Then gradient is: #(f(x+delta x)-f(x))/(x+delta x - x)# And the derivative is: #d/dx=lim_(delta x->0)((f(x+delta x)-f(x))/(x+delta x - x))# From example: #((-4(x+delta x)^2-5(x+delta x)-2)-(-4x^2-5x-2))/(x+delta x - x)# simplifying #(-4x^2-8xdelta x-4(deltax)^2-5x-5delta x-2+4x^2+5x+2)/(x+delta x - x)# #->=(-8xdelta x-4(deltax)^2-5delta x)/(delta x )# Cancelling #delta x# #(-8x-4(deltax)-5)# #d/dx=lim_(deltax->0)(-8x-4(deltax)-5)=-8x-5#

Charles Arocha · Answer

undefined To use the limit definition of the derivative to find the derivative of $ f(x) = -4x^2 - 5x - 2 $, we apply the following formula:

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

Substitute the given function into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{(-4(x + h)^2 - 5(x + h) - 2) - (-4x^2 - 5x - 2)}{h} $$

Expand and simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{-4(x^2 + 2xh + h^2) - 5x - 5h - 2 + 4x^2 + 5x + 2}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-4x^2 - 8xh - 4h^2 - 5x - 5h - 2 + 4x^2 + 5x + 2}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-8xh - 4h^2 - 5h}{h} $$
$$ f'(x) = \lim_{h \to 0} (-8x - 4h - 5) $$

Now, take the limit as $ h $ approaches 0:

$$ f'(x) = -8x - 5 $$

So, the derivative of $ f(x) = -4x^2 - 5x - 2 $ is $ f'(x) = -8x - 5 $.