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

Question

Isaiah Allen · Accepted Answer

#f(x)=1-x^2 => f'(x)=-2x#

In order to find a derivative with limits you need the definition.

#f'(x):=lim_(hto0)(f(x+h)-f(x))/h#

Basically the derivative of a function is the limit of the change of the function, as that change approaches 0. This is one reason why derivatives are used for things like "instantaneous rate of change."

In this case #f(x)=1-x^2# then #f(x+h)=1-(x+h)^2#

and #(x+h)^2=(x+h)(x+h)#

so by using FOIL we get

#(x+h)^2=(x+h)(x+h)=x x+xh+hx+hh#

#=x^2+2hx+h^2#

then

#f(x+h)=1-(x^2+2hx+h^2)=1-x^2-2hx-h^2#

So, we can plug this in

#f'(x)=lim_(hto0)(1-x^2-2hx-h^2-(1-x^2))/h#

Since #(-1)(1-x^2)=-1+x^2# we can say

#f'(x)=lim_(hto0)(1-x^2-2hx-h^2-1+x^2)/h#

Then we group like terms

#f'(x)=lim_(hto0)((1-1)+(-x^2+x^2)-2hx-h^2)/h#

#f'(x)=lim_(hto0)(0+0-2hx-h^2)/h=lim_(hto0)(-2hx-h^2)/h#

Then the h terms cancel

#f'(x)=lim_(hto0)(-2cancelhx-h^(cancel2))/cancelh=lim_(hto0)-2x-h#

This type of algebraic manipulation is very common when using the limit definition for derivation.

Since the limit of a sum is the sum of limits.

#lim_(hto0)-2x-h=-2x-lim_(hto0)h=-2x-0=underline(-2x)#

Emma Aldridge · Answer

undefined To find the derivative of $f(x) = 1 - x^2$ using limits, follow these steps:

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

2. Substitute $f(x) = 1 - x^2$ into the formula:
   $$f'(x) = \lim_{h \to 0} \frac{(1 - (x + h)^2) - (1 - x^2)}{h}$$

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

4. Evaluate the limit as $h$ approaches 0:
   $$f'(x) = -2x$$

Therefore, the derivative of $f(x) = 1 - x^2$ with respect to $x$ is $f'(x) = -2x$.