Question

y= x^2 + 4x +3/ sqrt x

Answer 1

What do you need to do with that?

Answer 2

oh sorry differentiate the function

Answer 3

Ok, do you remember the differentiation rules?

Answer 4

Yeah I know the quotient rule.

Answer 5

Great. As that is a sum (the easier case), you can split it intro 3 parts and differentiate them separately.

Answer 6

I got to here and was stuck. I was shown how to get the answer, but I dont understand how they did part of it.

Answer 7

heres what I have, (sqrtx)(2x+4)-(x^2+4+3) 1/2x^-1/2

Answer 8

over (sqrtx)^2

Answer 9

those square roots are what is confusing me
I dont know what to do with them.

Answer 10

i changed the one on the far right hand side to 1/2 x^-1/2 so I could use the derivative

Answer 11

To make it easier you can think of them as (something)^1/2. Then you can differentiatie it as if it was a power.

Answer 12

right. but after I do that im lost

Answer 13

When you have to differentiate\[3. (x)^{\frac{-1}{2}}\]?

Answer 14

That is where you get lost?

Answer 15

well yes. because that last square root you have to take the derivative. but after that, I dont know what to do. or at least the explanation in the book doesnt make any sense.

Answer 16

i wish i could take a picture of the answer the book gives and show you so you can explain it

Answer 17

It would be:\[3 . \frac{-1}{2} . (x)^\frac{-3}{2} . x \prime\]
Which you can rewrite as
\[\frac{-3}{2} . \frac{ 1 }{ \sqrt[3]{x^2}}\]

Answer 18

ok can you write this whole thing out? i apologize but I need to see the whole thing worked through to get it.

Answer 19

Sure. \[y = x^2 + 4x + \frac{3}{\sqrt{x}}\] So \[y \prime = (x^2)\prime + (4x)\prime + (\frac{3}{\sqrt{x}})\prime\] Because it is a sum, right?

Answer 20

before you get to far into this, you know there is a sqrt x in the denominator, right?

Answer 21

Oh, like this? \[y = \frac{x^2 + 4x + 3}{\sqrt{x}}\]

Answer 22

correct. I apply the quotient rule, and then I get stuck

Answer 23

\[y \prime = \frac{(x^2+4x+3)\prime . \sqrt{x} - (x^2+4x+3) . \sqrt{x}\prime }{\sqrt{x}^2}\]

Answer 24

good so far. after this is when I get wonky
\

Answer 25

So far you're good, right?

Answer 26

correct

Answer 27

Ok, let's take it by pieces so it's easier for you to see.

Answer 28

The sum is a nice one to differenciate, because you can split it into even smaller parts.
\[(x^2+3x+3)\prime = 2x + 3 + 0\]

Answer 29

You're good with that one?

Answer 30

yup

Answer 31

you mean 4x right?

Answer 32

Cool. Now let's go with the nasty square root.

Answer 33

Ups, yeah. My bad there!

Answer 34

no worries please continue

Answer 35

Ok, so the little trick with the square roots, or any root, is to think of them as a power, right?

Answer 36

Because you already have a rule to differentiate X to some power, and it's relatively easy...

Answer 37

x^1/2 right

Answer 38

Right. So applying the same method you get:
\[y \prime = (x^\frac{-1}{2})\prime = \frac{-1}{2} . X^{\frac{1}{2} - 1}\]

Answer 39

1/2x^-1/2 right

Answer 40

That is the same: \[y \prime = \frac{-1}{2} X^{\frac{-3}{2}}\]

Answer 41

Yep, there's a minus missing.

Answer 42

I'm awful with this editor.

Answer 43

wait why is it -1/2?

Answer 44

if sqrtx is x^-1/2, how does it become -1/2x^-1/2?

Answer 45

I made a mistake again. xD You're totally right

Answer 46

It's a positive 1/2

Answer 47

well at least im paying attention. so its 1/2x^-1/2 right?

Answer 48

\[y \prime = \frac{1}{2}X^{\frac{-1}{2}}\]

Answer 49

ok. after this is where im stuck

Answer 50

Now you can leave as it is, or you can change it back into a square root. And then plug it into the big quotient rule.

Answer 51

\[y \prime = \frac{1}{2} \frac{1}{\sqrt{x}} = \frac{1}{2 \sqrt{x}}\]

Answer 52

That's all the differentiation done. You just have to plug it in and make it look nice. :P

Answer 53

so its sqrtx(2x+4)-(x^2+4x+3)-1/2sqrtx ?

Answer 54

over sqrtx^2?

Answer 55

\[y \prime = \frac{(2x+4) . \sqrt{x} - (x^2 + 4x + 3) . \frac{1}{2 \sqrt{x}}} {x}\]

Answer 56

Exactly

Answer 57

wait so the sqrtx and the sqrtx^2 cancel?

Answer 58

\[y \prime = \frac{(2x+4)\sqrt{x} - {(x^2 + 4x + 3)}}{x 2\sqrt{x}}\]

Answer 59

Yes. Actually, you should put a module there.

Answer 60

im still lost. where did the numerator 1/2sqrtx go?

Answer 61

\[y \prime = \frac{(2x+4)\sqrt{x} - (x^2 + 4x +3)}{|x| 2\sqrt{x}}\]

Answer 62

To the denominator.

Answer 63

so they canceled?

Answer 64

No, it's stil there.

Answer 65

When you multiply by 1 over something, it is the same as dividing by that something.

Answer 66

So you can just put it in the denominator and it won't affect the result.

Answer 67

oh right

Answer 68

I hope I didn't confuse you more than you were. With my typos and all...

Answer 69

so what happened to the sqrtx^2 in the denominator?

Answer 70

You could leave it like that if you want to. But also you can simplify it by writing it as module of X.

Answer 71

If you take any number, square it, and then take the square root it is the same as if you just grabbed that number and made it always positive.

Answer 72

If it's easier for you, you can always leave it as the square root squared. It's not wrong.

Answer 73

ok. what does it look like now?

Answer 74

Check above, the last formula. That would be the final result.

Answer 75

Do you get it or you still have doubts?

Answer 76

well its not the right answer so yeah I still am confused. dont worry about it. im just going to ask the professor monday.

Answer 77

What does the answer say?

Answer 78

Is it much different?

Answer 79

3x^2+4x-3/2x^3/2

Answer 80

Oh, that's probably the result after distributing the square root of X in the numerator and then doing the division.

Answer 81

In the denominator if you multiply the 2 square root of X squared by X you get the 2 . (x)^3/2

Answer 82

yeah im still lost. im just going to move on.

Answer 83

Yeah, take a look at it later. But it's probably a different notation, if you got the quotient rule then you could do it just fine.

Answer 84

im not so sure. my algebra is pretty weak.

Answer 85

Don't worry, you'll get it. It's just a matter of practice. Then it becomes pretty mechanical.

Answer 86

well this is my third time taking calculus and its still hard as hell. maybe its the adhd.

Answer 87

I totally get it, I took it twice, too. But perseverance is everything in this stuff. Don't give up.

Answer 88

wasnt planning on it. just frustrated

Answer 89

getting tired of studying for hours on end and getting crappy grades

Answer 90

Right. It sucks.

Answer 91

But sometimes it comes a time where you magically start understanding.

Answer 92

well anytime would be great. im wasting money here.

Answer 93

And you are like "Wow! I can't believe it was that easy"