Answer:
\(f'(x) = 2x-2\)
Step-by-step explanation:
We are finding the derivative of the function:
\(f(x)=x^2-2x\)
using the definition of the derivative:
\(\displaystyle f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\)
Plugging in the given function, we get:
\(\displaystyle f'(x) = \lim_{h\to0}\dfrac{[(x+h)^2-2(x+h)]-(x^2-2x)}{h}\)
↓ expanding the numerator
\(\displaystyle f'(x) = \lim_{h\to0}\dfrac{[
ot\!x^2+2hx + h^2-
ot\!\!2x-2h]-
ot\!x^2+
ot\!\!2x}{h}\)
↓ canceling opposite terms
\(\displaystyle f'(x) = \lim_{h\to0}\dfrac{h^2 + 2hx - 2h}{h}\)
↓ factoring out an h from the numerator
\(\displaystyle f'(x) = \lim_{h\to0}\dfrac{
ot \!h(h + 2x - 2)}{
ot\!h}\)
↓ canceling the h in the numerator and denominator
\(\displaystyle f'(x) = \lim_{h\to0}(h+2x-2)\)
↓ evaluating the limit
\(\displaystyle f'(x) = 0+2x-2\)
\(\displaystyle \boxed{f'(x) = 2x-2}\)
