## ChrisV 3 years ago Find the derivative of the funtion. f(x)=x^2(x-2)^4

1. ChrisV

\[f(x)=x ^{2}(x-2)^{4}\]

2. .Sam.

Use product rule, y' = (x^2)(4)(x-2)^3+(x-2)^4 (2x) \[4x^{2}(x-2)^{3}+2x(x-2)^{4}\]

3. .Sam.

Teach you one technique, y=(left)(right) y'=(copy left)(differentiate right)+(copy right)(differentiate left)

4. ChrisV

yea product rule

5. ChrisV

you get 2x(x-2)^4+x^2(4(x-2)^3)

6. ChrisV

then apply the chain rule which i do not understand

7. .Sam.

Noneed to use chain rule here.

8. ChrisV

the answer in the back of the book says

9. ChrisV

\[x ^{2}[4(x-2)^{3}(1)] + (x-2)^{4}(2x) = 2x(x-2)^{3}(3x-2)\]

10. ChrisV

i dont understand hgow to get to 2x(x-2)^3(3x-2)

11. cuddlepony

f(x)=x^2(x-2)^4 split these things up durp, g(x) = x^(2) g'(x) = 2x d(x) = (x-2)^4 USE CHAIN RULE!!! s(x) = x^4 s'(x) = 4x^(3) j(x) = x-2 j'(x) = 1 so d'(x) =4(x-2)^(3)*1 so Now product rule g(x) = x^(2) g'(x) = 2x d(x) = (x-2)^4 d'(x) =4(x-2)^(3)*1 2x((x-2)^4) + (4(x-2)^(3))x^(2)

12. cuddlepony

easy

13. cuddlepony

f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2) Best way to deal with derivatives when you are first learning them is to split them into different functions and derive them then put them all together

14. cuddlepony

After you do this about a 1,000 times you will be able to do it in your head

15. cuddlepony

Any questions? also if you do it in the order I do it you wont mess up using Quotient rule

16. ChrisV

sorry let me read what you wrote

17. ChrisV

18. cuddlepony

Remember Chain rule is just g(x) = x^4 g'(x) = 4x^(3) s(x) = x-2 s'(x) = 1 we throw away g(x) and sub s(x) into g'(x) and multiply by s(x) derivative so g'(s(x)) * s'(x)

19. cuddlepony

this is the fool proof way of doing derivatives when you first start out at least it was for me

20. ChrisV

ok let me try to reword this

21. ChrisV

x^2(x+2)^4

22. Chlorophyll

f(x)=x^2(x-2)^4 f'(x) = 2x *(x-2)^4 + 4x^2 (x -2) ^3 = (x - 3) ^3 [ 2x(x-2) + 4x^2 ] = (x - 3) ^3 [ 6x^2 - 4x ]

23. ChrisV

(2x)((x+2)^4) + 4x^2(x-2)^3

24. cuddlepony

g(x)d(x) Take derivative of g(x) take derivative of d(x) you have to use chain rule to do so g'(x)d(x) + g(x)d'(x) if you do it in this order it works for quient rule as well g(x)/d(x) (g'(x)(d(x)) - g(x)d'(x))/(d(x))^(2)

25. ChrisV

ah so i then take (2x)((x+2)^4) + 4x^2(x-2)^3 and split it like [2x(x+2)^4 +4x^2](x-3)^3 ?

26. ChrisV

and do the product rule again?

27. cuddlepony

no!

28. cuddlepony

f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2) That is it you can simplify it but that is the derivative

29. cuddlepony

if you wanted to take the second derivative you would use product rule again.

30. ChrisV

ok so i need to simplify f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)

31. cuddlepony

its only asking for the first derivative so yeah. Is there anything you do not understand about the method I showed you? If you have to you can

32. ChrisV

so i factor out (x-2)^3

33. cuddlepony

you can also factor out 2x

34. cuddlepony

2x(x-2)^(3)(x - 2 + 2x)

35. ChrisV

so i would end up with 2x(x-2)^3(x-2+2x)

36. cuddlepony

2x(3x-2)(x-2)^(3)

37. cuddlepony

2x^(1) and x^(1) can be added together

38. ChrisV

you have been so much help

39. ChrisV

i had the right answer the whole time just didnt see the factor

40. cuddlepony

because they are to the same power, remember they have to be to the same power to add them you can't add x^(2) + x^(1) or x^(3) + x Not to be partrionizing

41. cuddlepony

*patronizing

42. cuddlepony

yeah math can be a hassel sometimes lol

43. ChrisV

no i understand now

44. ChrisV

thanks a lot

45. cuddlepony

np take it easy