## mathcalculus one year ago can someone help me find the derivative of a square root step by step please.

1. mathcalculus

Sqrt(2-5*x^2)

2. mathcalculus

i did this: 2-5x^2) ^1/2

3. mathcalculus

then 1/2(2-5x^2)

4. mathcalculus

@zepdrix

5. zepdrix

Woops you didn't apply the power rule correctly.$\large (2-5x^2)^{1/2}$Derivative gives us,$\large \frac{1}{2}(2-5x^2)^{1/2-1} \qquad=\qquad \frac{1}{2}(2-5x^2)^{-1/2}$But we also have to apply the chain rule after that, multiplying by the derivative of the inside.

6. mathcalculus

uh-oh got me there... may i ask how?

7. mathcalculus

i think i kind of do but im not sure how to proceed it.

8. zepdrix

$\large \color{royalblue}{\left[(2-5x^2)^{1/2}\right]'} \qquad = \qquad \frac{1}{2}(2-5x^2)^{-1/2}\color{royalblue}{(2-5x^2)'}$We took the derivative of the outer function, apply the power rule. Now we have to multiply by the derivative of the inside. The prime is to show that we still need to take the derivative of that part. Taking that derivative gives us,$\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)}$

9. zepdrix

Confused? :O Too much stuff going on?

10. mathcalculus

im a little confused with how you got to 10x

11. zepdrix

It's the derivative of that blue thing. Take the derivative of each term inside the brackets, separately.

12. mathcalculus

isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

13. mathcalculus

i see but what happens to the half?

14. zepdrix

That would be the product rule:$\large \left[f(x) \cdot g(x)\right]' \qquad=\qquad f'(x)g(x)+f(x)g'(x)$ We're dealing with a composition of functions, a function within a function. Which tells us to apply the chain rule:$\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)$

15. zepdrix

The chain rule tells us... After you've taken care of the outer function... (In our case our outer function is "something" to the 1/2 power) make a copy of the inside, take it's derivative, and multiply by it.

16. zepdrix

I hate using function notation like this, but maybe it'll help you to understand the chain rule. I find it more confusing, but to each his own. So we have this,$\large f(x)=x^{1/2}$$\large g(x)=2-5x^2$ $\large f(g(x))=(2-5x^2)^{1/2}$ Taking the derivative using the chain rule:$\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)$ $\large =\left[(2-5x^2)^{1/2}\right]'\left[2-5x^2\right]'$

17. zepdrix

|dw:1375768509806:dw|If we have a function of a potato... The chain rule tells us that after we take the derivative of the function, we make a copy of the potato and take it's derivative as well.

18. zepdrix

It's a tricky concept to get used to :(

19. Psymon

Haha

20. Psymon

I like that one :P

21. zepdrix

Uh oh, your brain esplode? :( Want to see a nice easy example? One which includes no potatoes? :o

22. mathcalculus

i understand the -10x but what happens to the -1/2 exponent.

23. mathcalculus

lol

24. Psymon

Next time we should do f ' (wal-mart)

25. zepdrix

We're making a copy of the inside and taking it's derivative. The -1/2 is on the outside.

26. zepdrix

We're not making a copy of the outside, if that makes sense :o

27. mathcalculus

i know the chain rule but i dont understand the other one

28. mathcalculus

after you find -10x then what

29. zepdrix

We have this delicious Chicken breast seasoned with rosemary and all this good stuff. Then we stuff it with cheese and maybe even potatoes, cause that's how we roll. When we take the derivative of this chicken, the chain rule tells us to make a copy of the cheese and potatoes. This copy will not include any of the rosemary or seasoning from the initial chicken. After the -10x? We're done!

30. zepdrix

That last example was so ... unnecessary lol... I should delete that.

31. Psymon

Haha, we need the humor with these things sometimes, though : )

32. mathcalculus

noooo dont lol

33. mathcalculus

don't delete :) thank you

34. mathcalculus

now, the critical points are= x=10?

35. mathcalculus

i mean for x

36. mathcalculus

i know how to solve for y

37. zepdrix

Oh since addition is commutative, we can rearrange things a bit just to make it look prettier.$\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)} \qquad=\qquad \frac{-5x}{\sqrt{2-5x^2}}$

38. zepdrix

Since multiplication is commutative* Now we need to find critical points?

39. zepdrix

I kinda threw a bunch of steps in there when I simplified. Did any of that confuse you? :o

40. mathcalculus

the chain rule, yes

41. mathcalculus

oh simplied? uhhh no not really. i know how you did that.

42. mathcalculus

@zepdrix

43. mathcalculus

oh wait i understand th chain rule now. simply find the derivative of the "inner" which to me... is 2-5x^2

44. zepdrix

yah that's the inner thing :o take it's derivative, and multiply by it.

45. mathcalculus

rewrite and first and multiply by the inner's derivative

46. mathcalculus

okay i caught up with ya.

47. mathcalculus

now to find that x.....

48. mathcalculus

because they;re asking for the critical points.

49. mathcalculus

@zepdrix

50. mathcalculus

@Psymon

51. Psymon

I didn't want to interfere xD

52. mathcalculus

it's okay, team work =]

53. mathcalculus

@Psymon

54. zepdrix

To find critical points we set the derivative function equal to zero. Critical points exist anywhere the derivative is zero, or does not exist.$\large 0=\frac{-5x}{\sqrt{2-5x^2}}$ So we can determine that a critical point exists when:$\large 0=-5x \qquad\qquad\qquad 0=\sqrt{2-5x^2}$

55. mathcalculus

x=5?

56. mathcalculus

those are not critical points..... :/ it should be written like this.. ex: (0,0)

57. zepdrix

From that first zero? Hmm it looks like if we divide both sides by -5, we get x=0.

58. zepdrix

I'm not sure where you're getting 5 from :o

59. zepdrix

No, normally you write critical points as x= Max and min are written as coordinate pairs. Maybe your teacher does it differently though :)

60. mathcalculus

oops! right x=0

61. zepdrix

How bout this: $\large 0=\sqrt{2-5x^2}$ What critical points is it giving us?

62. Psymon

Yeah, I write crit points as x = as well, only a coordinate pair for min and max. Just to kinda back up zep, lol.

63. mathcalculus

does not exist?

64. zepdrix

Squaring both sides gives us,$\large 0=2-5x^2$Solving for x gives us,$\large x=\pm \sqrt{2/5}$

65. mathcalculus

wait a min i had that... :/ but it was x= sqrt(2/-5)

66. mathcalculus

because the - is there^..

67. zepdrix

Hmm you missed a negative somewhere :o Subtracting 2 from each side gives us -2 on the left. The dividing by -5 gives us positive 2/5 on the left.

68. mathcalculus

oops, right again.

69. mathcalculus

@ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.

70. zepdrix

np c: cya