Matthew071 2 years ago Given that f(x)=-2h(x)+(2x/h(x)); h(4)=4, h'(4)=-2. Find f'(4)

1. Matthew071

well their is: a)13/2 b)7/2 c)-5/2 d)11/2 e)17/2

2. Matthew071

I tried solving for the derivative of f(x) but I must be doing something wrong or the awnser is not there

3. myininaya

So what did you get for f'?

4. myininaya

I will check it.

5. Matthew071

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

6. myininaya

Also the answer is there. h(x) shouldn't be there. you are missing h' in your quotient

7. myininaya

(-2h)'=-2(h)'=-2h'

8. Matthew071

oh ok

9. myininaya

(2x/h)'=[(2x)'h-2x(h)']/h^2

10. Matthew071

so the awnser should be 19/2

11. myininaya

hmm... that isn't what I got...let me check my result.

12. myininaya

Oh you are still assuming h is there aren't you?

13. myininaya

(-2h)'=-2(h)'=-2h' not h-2h'

14. Matthew071

So how does it apply

15. myininaya

what do you mean?

16. Matthew071

well I am plugging in everything where $h(4)+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = 4 +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)$

17. Matthew071

correct?

18. myininaya

Well see I keep telling you (-2h)'=-2h' not h-2h'

19. myininaya

That first term you are putting in shouldn't be there.

20. Matthew071

Could you show me what you plugged in so I understand

21. myininaya

So you don't understand why (-2h)' is -2h' and not h-2h'?

22. Matthew071

No

23. myininaya

I'm liking I'm using the constant multiple rule.

24. myininaya

(cf)'=cf' Let k(x)=cf(x) To find k'(x) we use the formal definition of derivative: $k'(x)=\lim_{\Delta x \rightarrow 0} \frac{k(x+\Delta x)-k(x)}{\Delta x}=\lim_{\Delta x \rightarrow 0} \frac{cf(x+\Delta x)-cf(x)}{\Delta x}$ Note: By the distributive property ab+ac=a(b+c) $=\lim_{\Delta x \rightarrow 0} c \frac{f(x+\Delta x)-f(x)}{\Delta x}$ Note: By limit properties, we can do lim c g(x) = c lim g(x) $=c \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$ Note: By definition of derivative that above is cf'(x).

25. myininaya

The constant multiple rule: k=cf k'=cf'

26. myininaya

So say k=ch then k'=ch' That c means constant The constant in front of h in your problem is -2 so k=-2h then k'=-2h'

27. Matthew071

So when I plug in for -2h' when h(4)=4 and h'(4)=-2

28. myininaya

Yep

29. myininaya

-2h'(x) x is 4. -2h'(4) -2h'(4)=-2(-2)=4 Now simplify your quotient and add.

30. Matthew071

Does this mean -2h'= -2(-2) or -2(4)

31. Matthew071

ok

32. myininaya

h' is h'(x)

33. myininaya

[-2h(x)] ' =-2 [ h(x)] ' =-2 h'(x) This has been what I have been saying but you wrote it as h(x)-2h'(x)

34. myininaya

$\cancel{h(4)}+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = \cancel{4} +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)$

35. myininaya

This is what you wrote earlier I'm telling you (-2h)'=-2h' not h-2h' do you understand now?

36. Matthew071

Oh ok Now I get what you mean by this

37. myininaya

I put the marks through the bad parts.

38. myininaya

so you understand the constant multiple rule?

39. Matthew071

Yes I do now understand the multiple rule and the awnser is 11/2

40. myininaya

That is right.

41. Matthew071

I just didnt understand it conceptually but now that I see it plugged in and used I do now

42. myininaya

Ok. Neatness.

43. Matthew071

Thank you so much

44. myininaya

Np.