Math Problem: (Don't know how to solve this. Please explain when showing how to solve it.)

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Math Problem: (Don't know how to solve this. Please explain when showing how to solve it.)

Mathematics
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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 \[\frac{ d }{ dx }x = x ^{n-1}\]

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hmm first you need to find f(x+h) so substitute x for (x+h) in f(x) function \[\huge\rm f(x+h)=\color{reD}{x}^2-2\color{ReD}{x}\]
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 :(
\[\huge\rm f(x+h)=\color{reD}{(x+h)}^2-2\color{ReD}{(x+h)}\] solve that use foil method for (x+h)^2 and distribute parentheses by -2
ok...power rule is like (ill just show for x^2) \[\frac{ d }{ dx }(x^2) = 2x\]
How did you get 2x? what was the value of d @saseal?
actually they call it dy/dx it means to differentiate
So the answer for that problem would be\[(x ^{2} +2xh +h ^{2})-2x-2h\] @Nnesha?
so are you suppose to plug in y and x instead of d @saseal?
no not the answer that's just f(x+h) now use that formula \[\huge\rm \frac{ f(x+h)-f(x) }{ h }\] replace f(x+h) by (x^2 +2xh+h^2)-2x-2h and f(x) by the function x^2-2x
\[\huge\rm \frac{ \color{ReD}{f(x+h)}-\color{blue}{f(x)} }{ h }\] \[\huge\rm \frac{ \color{reD}{ x^2 +2xh+h^2-2x-2h } -\color{blue}{(x^2 -2x) }}{ h }\] simplify !
\[f'(x)=\frac{((x+h)^2-2x+h)- (x^2-2x)}{ h}\]
(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?
\[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\] \[\large\rm \frac{ h^2+2xh+h }{ h}\] \[\large\rm \frac{ h(h+2x+1)}{ h}\]
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
\[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\] \[\large\rm \frac{ h^2+2xh\color{ReD}{-2h} }{ h}\] \[\large\rm \frac{ h(h+2x\color{reD}{-2})}{ h}\] i forgot it suppose to be -2h not just h
thanks saseal
\[\large\rm \frac{ \cancel{h}(h+2x\color{reD}{-2})}{\cancel{ h}}\]
Ok, as promised the power rule: you take the power of the variable and pull it down and minus 1 to the power like this \[(\frac{ dy }{ dx })2x^2 = (2)2x ^{2-1}\]
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.

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