Using the limit definition, how do you differentiate #f(x)=1-x^2#?

Question

William Abdalla · Accepted Answer

If #f(x) = 1-x^2# then #(d f(x))/(dx) = -2x# The limit definition of the derivative of a function #f(x)# is #color(white)("XXX")(d f(x))/(dx) = lim_(hrarr0) (f(x+h)-f(x))/h# If #f(x) = 1-x^2# #color(white)("XXX")#then (by the limit definition) #(d f(x))/(dx)# #color(white)("XXX")=lim_(hrarr0) ((1-(x+h)^2)-(1-x^2))/h# #color(white)("XXX")= lim_(hrarr0) (1-x^2-2xh-h^2-1+x^2)/h# #color(white)("XXX")=lim_(hrarr0)(-2xh-h^2)/h# #color(white)("XXX")=lim_(hrarr0) -2x-h# #color(white)("XXX")=-2x#

Scarlett Allison · Answer

undefined To differentiate the function $f(x) = 1 - x^2$ using the limit definition of a derivative, we apply the following formula:

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

Substitute the function $f(x) = 1 - x^2$ into the formula:

$$f'(x) = \lim_{h \to 0} \frac{(1 - (x + h)^2) - (1 - x^2)}{h}$$

Expand and simplify:

$$f'(x) = \lim_{h \to 0} \frac{1 - (x^2 + 2xh + h^2) - 1 + x^2}{h}$$

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

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

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

Now, as $h$ approaches 0, the limit simplifies to:

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

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