## anonymous 4 years ago need help with the attachment

1. anonymous

2. anonymous

a) Just plug in 1 for the numerator and g(x) for the denominator in the quotient rule. b) $-{\frac {1+k{{\rm e}^{s}}}{ \left( s+k{{\rm e}^{s}} \right) ^{2}}}$ assuming y is a function of s. c) Follows from $x^{-n} = \frac{1}{x^n}$ and $\frac{d}{dx} x^n = nx^{n-1}$

3. anonymous

is the quotient rule the same thing as the reciprocal part b?

4. anonymous

What do you mean? The reciprocal rule is a special case of the quotient rule. You can use either one for part b, but it's easier to use the reciprocal rule.

5. anonymous

The quotient rule f'g-fg'/g^(2)

6. anonymous

but g(x) is just a variable

7. anonymous

ok I got it

8. anonymous

What is your question? g is not just a variable, it's a function.

9. anonymous

10. anonymous

OK I need help with differentiating the bottom on part could you show me step by step how to differentiate that one?

11. anonymous

part b

12. anonymous

and also the numerator of the function

13. anonymous

You need to differentiate$\Large s+k\cdot e^s$ with respect to s.

14. anonymous

The derivative of a sum is the sum of derivatives, so differentiate s and then differentiate k*e^s.

15. anonymous

What is the derivative of s?

16. anonymous

17. anonymous

1

18. anonymous

Yes. And what is the derivative of e^s ?

19. anonymous

@ta, see the attachee

20. anonymous

e^(s)

21. anonymous

22. anonymous

Yes, so now you can add them together and you get the derivative

23. anonymous

And then just plug that into the formula in part a)

24. anonymous

1+e^(s)

25. anonymous

No! Don't forget the factor k!

26. anonymous

k does not depend on s, so you can just leave it unchanged

27. anonymous

But you must not forget it :D

28. anonymous

29. anonymous

here u r

30. anonymous

raheen there is still no explanation of part b :D

31. anonymous

32. anonymous

part b asking to find using the rule only

33. anonymous

ta123 do you have more questions?

34. anonymous

whears the other 2 parts are to prove

35. anonymous

@ta got that?

36. anonymous

lol raheen are you just trying to get a medal, because I can give you one if thats what you want :D

37. anonymous

38. anonymous

no YThe, I always try to help , pls don read the things in wrong way

39. anonymous

40. anonymous

yes, thanks for the help Y and raheen

41. anonymous

ta why some ppl are using an offensive words?

42. anonymous

I think I stayed with u b4 more than 2 hrs, without any annyance, right?

43. anonymous

yes

44. anonymous

thank u.

45. anonymous

no prob!