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Evaluate\[(f(x+h)-f(x))/h\] and simplify if \[f(x) = x^2-2x\]

this stuff is differentiation

i forgot how to do it this method...but i normally use power rule
\[\frac{ d }{ dx }x = x ^{n-1}\]

so i need to solve for x using that equation u gave me nnesha?

and i dont know that power rule saseal

not for x simplify that
that's the equation i gave you
you already have the formula

and i don't know that either :(

ok...power rule is like (ill just show for x^2)
\[\frac{ d }{ dx }(x^2) = 2x\]

actually they call it dy/dx it means to differentiate

\[f'(x)=\frac{((x+h)^2-2x+h)- (x^2-2x)}{ h}\]

(2xh+h^2-2h)/h

That's what I got after putting like terms together

ill show you the power rule magic after you get everything together

there is a common factor h^2+2xh+h

Alright

Oh ya. h(h+2x+1)

Is that the final answer?

no

Am I suppose to factor or insert it into another equation?

can you cancel out h?

Oh whoops I forgot about the h on the bottom. So the answer would actually be h+2x+1?

yep

Alright thanks nnesha and saseal!

where did the coefficient of -2h go?

hmm let's see.

I think we wrote that step wrong.

Alright its 2x-2+h

thanks saseal

\[\large\rm \frac{ \cancel{h}(h+2x\color{reD}{-2})}{\cancel{ h}}\]

Okay

if its just 2x it becomes
\[\frac{ dy }{ dx }(2x ^{1}) = 2x^0=2(1)\]

why did it become 2x^0?

so for your equation x^2-2x
\[\frac{ dy }{ dx }(x^2-2x) = 2x-2\]

because x^1 is x, after you differentiate it you need to -1 from its power so it becomes x^0

Okay I think I understand what you did. Would we learn this in AP Calculus AB?

if you wonder where the h went...its removed by limits as h approaches 0

Yea you would

it's like the meat of the calculus

Oh okay. Did you take AP Calculus?

No, AP is not available in my country, we do A levels instead which is about the same

i don't live in us

Oh okay sorry about that.