## mathcalculus Group Title can someone help me solve this problem ? let f(x)= (x^3+3x+8)^2 F'(x)= F'(2)= one year ago one year ago

1. hartnn Group Title

do you know the chain rule ?

2. mathcalculus Group Title

not really sure with the chain rule..

3. sam1194 Group Title

need to learn that then to do a question of this type

4. mathcalculus Group Title

yes, that's why I'm trying to learn it here.. if that helps.

5. hartnn Group Title

let me give you a similar example, if $$f(x) = (ax^2+bx+c)^3$$ then by chain rule, $$f'(x)=3(ax^2+bx+c)^2 (ax^2+bx+c)' \\ =3(ax^2+bx+c)^2(2ax+b)$$ got this ?

6. hartnn Group Title

by definition, $$f(g(x))' = f'(g(x)) \times g'(x)$$

7. mathcalculus Group Title

yeah

8. hartnn Group Title

so try once for your Q, maybe.. ?

9. mathcalculus Group Title

you dont subtract the exponent n-1?

10. hartnn Group Title

i made the exponent n-1 in the Q, ax^2+bx+c had exponent 3, in the answer i became 2, right ?

11. hartnn Group Title

for x^n, the derivative is $$(x^n)'=n x^{n-1}$$

12. mathcalculus Group Title

okay... sort of there. but then the second equation next to it confused me a little.

13. hartnn Group Title

yeah, from the definition, i needed to multiply by the derivative of inner function also (g'(x)) so ,i multiplied by derivative of ax^2+bx+c which is 2ax+b

14. hartnn Group Title

in your Q, you'll multiply by (x^3+3x+8)

15. hartnn Group Title

**by the derivative of (x^3+3x+8)

16. mathcalculus Group Title

so the derivative for F'(x) would be.. (3x^2+3x+8)?

17. hartnn Group Title

no... you have to first derivate the out function, (x^2), then multiply the derivative of each term of x^3+3x+8 first tell whats the derivative of x^2

18. mathcalculus Group Title

2x

19. hartnn Group Title

so, derivative of (x^3+3x+8)^2 will be 2(x^3+3x+8) *(derivative of x^3+3x+8 ) , right ? can you differentiate, x^3+3x+8 ?

20. mathcalculus Group Title

2x^2+6x+16?

21. hartnn Group Title

no ? how u got that ?

22. mathcalculus Group Title

multiplied...

23. hartnn Group Title

derivative of x^3+3x+8 lets take it term by term derivative of x^3 =... ?

24. mathcalculus Group Title

3x

25. hartnn Group Title

no... derivative of x^n = nx^{n-1}

26. mathcalculus Group Title

3x^2

27. hartnn Group Title

derivative of 3x ?

28. mathcalculus Group Title

3

29. hartnn Group Title

derivative of 8 ?

30. mathcalculus Group Title

2? i really am so confused

31. hartnn Group Title

derivative of a constant = 0 so derivative of 8 is 0 so, derivative of x^3+3x+8 = 3x^2+3 right ?

32. mathcalculus Group Title

oh okay.

33. mathcalculus Group Title

i understand that better.

34. mathcalculus Group Title

now what if 2 was plugged in

35. hartnn Group Title

so, we have : derivative of (x^3+3x+8)^2 will be 2(x^3+3x+8) *(derivative of x^3+3x+8 ) , = 2(x^3+3x+8) * (3x^2+3) got this ?

36. mathcalculus Group Title

i thought the derivative of f '(x) is 3x^2+3

37. hartnn Group Title

didn't i mention to u the chain rule ?

38. mathcalculus Group Title

oh yeah, multiple use the one before and after

39. hartnn Group Title

40. mathcalculus Group Title

nope no doubts. now what if the problem was f ' (2)= ?

41. hartnn Group Title

just plug in x=2

42. mathcalculus Group Title

i did but im still not sure what am i suppose to do.

43. mathcalculus Group Title

i plug them in but i get confused with the derivative and actually multiplying it out.

44. hartnn Group Title

2(x^3+3x+8) * (3x^2+3) put x=2 2(2^3+3*2+8) * (3*2^2+3) =... ?

45. mathcalculus Group Title

44

46. hartnn Group Title
47. mathcalculus Group Title

thank you. thats correct

48. hartnn Group Title

welcome ^_^

49. mathcalculus Group Title

would it be possible to know the answer of a few problems because I would need to sign off and practice by hand but I would like to check my work and see if they're correct.