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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Evaluate\[(f(x+h)f(x))/h\] and simplify if \[f(x) = x^22x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this stuff is differentiation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i forgot how to do it this method...but i normally use power rule \[\frac{ d }{ dx }x = x ^{n1}\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1hmm first you need to find f(x+h) so substitute x for (x+h) in f(x) function \[\huge\rm f(x+h)=\color{reD}{x}^22\color{ReD}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i need to solve for x using that equation u gave me nnesha?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i dont know that power rule saseal

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1not for x simplify that that's the equation i gave you you already have the formula

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1and i don't know that either :(

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm f(x+h)=\color{reD}{(x+h)}^22\color{ReD}{(x+h)}\] solve that use foil method for (x+h)^2 and distribute parentheses by 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok...power rule is like (ill just show for x^2) \[\frac{ d }{ dx }(x^2) = 2x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you get 2x? what was the value of d @saseal?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually they call it dy/dx it means to differentiate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer for that problem would be\[(x ^{2} +2xh +h ^{2})2x2h\] @Nnesha?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so are you suppose to plug in y and x instead of d @saseal?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1no 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)2x2h and f(x) by the function x^22x

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm \frac{ \color{ReD}{f(x+h)}\color{blue}{f(x)} }{ h }\] \[\huge\rm \frac{ \color{reD}{ x^2 +2xh+h^22x2h } \color{blue}{(x^2 2x) }}{ h }\] simplify !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f'(x)=\frac{((x+h)^22x+h) (x^22x)}{ h}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I got after putting like terms together

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ill show you the power rule magic after you get everything together

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1there is a common factor h^2+2xh+h

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the final answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Am I suppose to factor or insert it into another equation?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{2x}2h } \color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\] \[\large\rm \frac{ h^2+2xh+h }{ h}\] \[\large\rm \frac{ h(h+2x+1)}{ h}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh whoops I forgot about the h on the bottom. So the answer would actually be h+2x+1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright thanks nnesha and saseal!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did the coefficient of 2h go?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think we wrote that step wrong.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{2x}2h } \color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\] \[\large\rm \frac{ h^2+2xh\color{ReD}{2h} }{ h}\] \[\large\rm \frac{ h(h+2x\color{reD}{2})}{ h}\] i forgot it suppose to be 2h not just h

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \frac{ \cancel{h}(h+2x\color{reD}{2})}{\cancel{ h}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, as promised the power rule: you take the power of the variable and pull it down and minus 1 to the power like this \[(\frac{ dy }{ dx })2x^2 = (2)2x ^{21}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if its just 2x it becomes \[\frac{ dy }{ dx }(2x ^{1}) = 2x^0=2(1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why did it become 2x^0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for your equation x^22x \[\frac{ dy }{ dx }(x^22x) = 2x2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because x^1 is x, after you differentiate it you need to 1 from its power so it becomes x^0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I think I understand what you did. Would we learn this in AP Calculus AB?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you wonder where the h went...its removed by limits as h approaches 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's like the meat of the calculus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay. Did you take AP Calculus?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, AP is not available in my country, we do A levels instead which is about the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay sorry about that.
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