Differentiate y=x^2-x using first principles

Thank you. The answer is 2x?
You have [math]f(x) = x^2 - x[/math]
Don't just guess. Actually populate the formula provided above and SHOW that result. Then, use your best algebra and SHOW that result.
 
Thank you. The answer is 2x?
No. Did you find \(\displaystyle f(x+h)-f(x)\) correctly?
\(\displaystyle f(x+h)-f(x)=[(x+h)^2-(x+h)]-(x^2-x)\\=[(x^2+2xh+h^2)-(x+h)]-[(x^2-x)]\\=(2x-1)h+h^2\)
 
You have [math]f(x) = x^2 - x[/math]
Don't just guess. Actually populate the formula provided above and SHOW that result. Then, use your best algebra and SHOW that result.

The answer is 2x=1 ?? sorry I do not know how to type process and I can show process
 
You can try to use existing notation.

You have f'(x) on your hand-written version. Why not use it in the final result?

I presume "=" is a typo and you meant "-".
 
OK. You were close. Here is how I would do it in steps to reduce the chance of stupid error.

[MATH]f(x) = x^2 - x.[/MATH]
[MATH]\therefore f(x + h) = (x + h)^2 - (x + h) = x^2 + 2hx + h^2 - (x + h) = (x^2 - x) + 2hx + h^2 - h.[/MATH]
[MATH]\therefore f(x + h) - f(x) = 2hx + h^2 - h \implies[/MATH]
[MATH]\dfrac{f(x + h) - f(x)}{h} = \dfrac{2hx + h^2 - h}{h} = 2x - 1 + h \implies[/MATH]
[MATH]\lim_{h \rightarrow 0}\dfrac{f(x + h) - f(x)}{h} = ( \lim_{h \rightarrow 0}2x - 1 + h) = (\lim_{h \rightarrow 0} 2x - 1) + ( \lim_{h \rightarrow 0} h) = 2x - 1 + 0 = 2x - 1 \implies f'(x) = 2x - 1.[/MATH]
Calculate and simplify algebraically f(x + h) as a separate step. In the next step, subtract f(x) from the result. In the third step, divide by h. And finally take the limit. Now you can do this by simply implementing the definition as the limit of the Newton quotient right at the start, but I have found working up to that result in steps can avoid many errors.
 
Last edited:
Top