Answer:
[tex]f'(x) = 2x-2[/tex]
Step-by-step explanation:
We are finding the derivative of the function:
[tex]f(x)=x^2-2x[/tex]
using the definition of the derivative:
[tex]\displaystyle f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}[/tex]
Plugging in the given function, we get:
[tex]\displaystyle f'(x) = \lim_{h\to0}\dfrac{[(x+h)^2-2(x+h)]-(x^2-2x)}{h}[/tex]
↓ expanding the numerator
[tex]\displaystyle f'(x) = \lim_{h\to0}\dfrac{[\not\!x^2+2hx + h^2-\not\!\!2x-2h]-\not\!x^2+\not\!\!2x}{h}[/tex]
↓ canceling opposite terms
[tex]\displaystyle f'(x) = \lim_{h\to0}\dfrac{h^2 + 2hx - 2h}{h}[/tex]
↓ factoring out an h from the numerator
[tex]\displaystyle f'(x) = \lim_{h\to0}\dfrac{\not \!h(h + 2x - 2)}{\not\!h}[/tex]
↓ canceling the h in the numerator and denominator
[tex]\displaystyle f'(x) = \lim_{h\to0}(h+2x-2)[/tex]
↓ evaluating the limit
[tex]\displaystyle f'(x) = 0+2x-2[/tex]
[tex]\displaystyle \boxed{f'(x) = 2x-2}[/tex]
