## farukk 2 years ago ???

1. farukk

which is correct? i.e a+b\a-b (after rationallization)a+b\a-b * a+b\a+b or a+b\a-b * a-b\a-b ?

2. UnkleRhaukus

|dw:1354620267974:dw|

3. UnkleRhaukus

|dw:1354620294927:dw|

4. farukk

can it be a-b rather than a+b?

5. UnkleRhaukus

it would be equal but it wouldn't helpful in rationalisation

6. farukk

all right thanks

7. UnkleRhaukus

|dw:1354620486122:dw|

8. farukk

one more question ?

9. farukk

10. UnkleRhaukus

yes!

11. farukk

|dw:1354681791055:dw|

12. farukk

u got?

13. farukk

how will we use power rule in this question?

14. UnkleRhaukus

so the first step would be to remember that the the derivative of a sum , is the sum of derivatives

15. farukk

yes yes

16. UnkleRhaukus

$\frac{\text d}{\text d x}(1+\sqrt{1-x^2})=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)$

17. farukk

derivative of a constant will be 0

18. UnkleRhaukus

right

19. UnkleRhaukus

so the problem is much simpler now

20. UnkleRhaukus

can you remember the derivative of a function of a function?

21. farukk

to me it's start right now:)

22. farukk

i'm sorry

23. farukk

|dw:1354682465787:dw|

24. UnkleRhaukus

$(f\circ g)'(x)=f(g(x))'=f'(g(x))g'(x)$

25. UnkleRhaukus

|dw:1354621505691:dw|

26. UnkleRhaukus

you have to multiply by the derivative of the the inside function

27. UnkleRhaukus

|dw:1354621612577:dw|

28. UnkleRhaukus

following?

29. farukk

|dw:1354682900778:dw|

30. UnkleRhaukus

yeah

31. farukk

but how?

32. UnkleRhaukus

$\frac{\text d}{\text d x}(1+\sqrt{1-x^2})$$=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)$$=\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)$ $\qquad\frac{\text d}{\text dx}f(g(x))=(f\circ g)'(x)=g'(x)f'(g(x))$$\qquad f(x)= x^{1/2}\qquad f'(x)=\frac12x^{-1/2}$$\qquad g(x)=1-x^2\qquad g'(x)=-2x$ $=-2x\times\frac12(1-x^2)^{-1/2}$

33. UnkleRhaukus

i know this stuff is tricky, can you tell me where about you are having trouble

34. farukk

haha

35. farukk

it is but u have solved it thank u so much!!

36. UnkleRhaukus

yeah but do understand the chain rule/,

37. farukk

no will be tomorow in college

38. farukk

hey did you use chain rule?

39. UnkleRhaukus

the chain rule for the derivative of a function of a function $(f\circ g)'(x)=g'(x)f'(g(x))$

40. farukk

is there any other way for this question except chain rule?

41. farukk

u didt use power rule? did u?

42. UnkleRhaukus

i used the power fule to find f '(x) and g'((x)

43. farukk

confusing me

44. UnkleRhaukus

watch this video , skip the first 5 minutes http://www.khanacademy.org/math/calculus/differential-calculus/e/chain_rule_1

45. farukk

actually i have solved all my questions except this one in quite a simple way

46. UnkleRhaukus

did you get the right answers?

47. farukk

i'm solving questions using theorems

48. farukk

there is no video

49. farukk

50. UnkleRhaukus

well you will have to learn the chain rule

51. UnkleRhaukus
52. farukk

|dw:1354747831157:dw|

53. farukk

mistake* -2x\ 2 sqrt 1 - x square

54. farukk

|dw:1354748818882:dw|

55. UnkleRhaukus

you can cancel the two

56. farukk

yeah rightly said!