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

1. anonymous

Evaluate$(f(x+h)-f(x))/h$ and simplify if $f(x) = x^2-2x$

2. anonymous

this stuff is differentiation

3. anonymous

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

4. Nnesha

hmm first you need to find f(x+h) so substitute x for (x+h) in f(x) function $\huge\rm f(x+h)=\color{reD}{x}^2-2\color{ReD}{x}$

5. anonymous

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

6. anonymous

and i dont know that power rule saseal

7. Nnesha

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

8. Nnesha

and i don't know that either :(

9. Nnesha

$\huge\rm f(x+h)=\color{reD}{(x+h)}^2-2\color{ReD}{(x+h)}$ solve that use foil method for (x+h)^2 and distribute parentheses by -2

10. anonymous

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

11. anonymous

How did you get 2x? what was the value of d @saseal?

12. anonymous

actually they call it dy/dx it means to differentiate

13. anonymous

So the answer for that problem would be$(x ^{2} +2xh +h ^{2})-2x-2h$ @Nnesha?

14. anonymous

so are you suppose to plug in y and x instead of d @saseal?

15. Nnesha

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

16. Nnesha

$\huge\rm \frac{ \color{ReD}{f(x+h)}-\color{blue}{f(x)} }{ h }$ $\huge\rm \frac{ \color{reD}{ x^2 +2xh+h^2-2x-2h } -\color{blue}{(x^2 -2x) }}{ h }$ simplify !

17. anonymous

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

18. anonymous

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

19. anonymous

That's what I got after putting like terms together

20. anonymous

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

21. Nnesha

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

22. anonymous

Alright

23. anonymous

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

24. anonymous

25. anonymous

no

26. anonymous

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

27. Nnesha

$\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }$ $\large\rm \frac{ h^2+2xh+h }{ h}$ $\large\rm \frac{ h(h+2x+1)}{ h}$

28. Nnesha

can you cancel out h?

29. anonymous

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

30. Nnesha

yep

31. anonymous

Alright thanks nnesha and saseal!

32. anonymous

where did the coefficient of -2h go?

33. Nnesha

hmm let's see.

34. anonymous

I think we wrote that step wrong.

35. anonymous

Alright its 2x-2+h

36. Nnesha

$\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }$ $\large\rm \frac{ h^2+2xh\color{ReD}{-2h} }{ h}$ $\large\rm \frac{ h(h+2x\color{reD}{-2})}{ h}$ i forgot it suppose to be -2h not just h

37. Nnesha

thanks saseal

38. Nnesha

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

39. anonymous

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 $(\frac{ dy }{ dx })2x^2 = (2)2x ^{2-1}$

40. anonymous

Okay

41. anonymous

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

42. anonymous

why did it become 2x^0?

43. anonymous

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

44. anonymous

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

45. anonymous

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

46. anonymous

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

47. anonymous

Yea you would

48. anonymous

it's like the meat of the calculus

49. anonymous

Oh okay. Did you take AP Calculus?

50. anonymous

@saseal

51. anonymous

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

52. anonymous

i don't live in us

53. anonymous