Question

Calculate the Derivative. Picture attached.

Answer 1

Odd question, not really sure what to think about it. I can follow the process, but would like so guidance along the way.

Answer 2

The first part setup makes sense, However when they use the chain rule, i don't see how they get that.

Answer 3

I might have just made some sense of it, is it the product rule or quotient rule inside of the chain rule?

Answer 4

Yup. Quotient rule is why you see a + there.

Answer 5

Would you mind going through the steps, i can't get the same result on my paper.

Answer 6

Product rule:
\((\pi*r)' *\sqrt{r^2 + h^2} + \sqrt{r^2+h^2}' *(\pi*r)\)

Answer 7

\(\Large \sqrt{r^2+h^2} = (r^2+h^2)^{1/2}\)
So that derivative comes out to:
\(\Large \frac{1}{2}(r^2+h^2)^{-1/2} * (r^2+h^2)'\)
By chain rule

Answer 8

Wait i dont see why we use the chian rule there.

Answer 9

Whats the inner and outer function?

Answer 10

*facepalms* nevermind

Answer 11

Outer function is square root, which I'm treating as a power of (1/2) for convenience.

Answer 12

Continue my bad

Answer 13

Inner function is \((r^2+h^2)\)

Answer 14

I got it. Sorry go ahead with what you were saying.

Answer 15

Cool, okay so that comes out to
\(\Large \frac{2r*r' + 2h}{2\sqrt{r^2+h^2}}\)

Answer 16

I don't follow that.

Answer 17

Which part of it? \(\Large (r^2+h^2)^{-1/2}\) becomes \(\Large \frac{1}{\sqrt{r^2+h^2}}\)

Answer 18

Oh so the square root goes to the bottom?

Answer 19

Yes. That's how negative exponents work. It moves things from numerator to denominator and becomes a positive exponent.

Answer 20

So i get the denominator, but how does the numerator become all that. Wouldn't it just be 2?

Answer 21

Because me multiply it by 1/2 we had previously

Answer 22

Chain rule says it gets multiplied by the derivative of the inside function, which in this case was \((r^2+h^2)\)

Answer 23

So the derivative just gets rid of the square root of the inside function basically?

Answer 24

Then where does the extra r come from in the numerator? Also what happened to the rest of the terms from the product rule step we did earlier?

Answer 25

The extra terms I've just been ignoring for now =) They're easier to make sense of.
The extra r just comes from using implicit differentiation. Any time you're taking the derivative of r, just multiply by r'. It's chain rule.

Answer 26

Okay, i'm pretty confused now. You mind backing up? I understand the 1/(sqrt(r^2 + h^2) Thats it.

Answer 27

If you could show each step, i'd really appreciate it. I know all the rules, but i get them all mixed up when combining multiple. Order of operations messed me up for this.

Answer 28

Sure man.

Answer 29

Thanks

Answer 30

Taking the derivative of \(\Large \pi*r\sqrt{r^2+h^2}\)
Product rule gives:
\(\Large (\pi*r)'*\sqrt{r^2+h^2} + \sqrt{r^2+h^2}'*\pi*r\)

Answer 31

Dealing with the left term, derivative of pi*r is just pi, but we multiply by r' because of implicit differentiation
\(\Large \pi*r'*\sqrt{r^2+h^2} + \sqrt{r^2+h^2}'*\pi*r\)

Answer 32

I understand implicit differencation, but don't see how it applies here.

Answer 33

Why wouldn't we just take pi as the derivative, don't see why we need to add r'?

Answer 34

There are two variables, r and h, and we're taking the derivative with respect to h.

Answer 35

So r' is \(\Large \frac{\text{dr}}{\text{dh}}\)

Answer 36

Derivative of r is not just 1, since we're deriving with respect to h, not with respect to r.

Answer 37

Okay, but that still doesn't explain why the derivative of pir is pi *r'

Answer 38

Oh okay so thats how it always works, with multiple variables?

Answer 39

Yes. You have to be conscious of what you are deriving with respect TO. Notice elsewhere in the problem when we take derivative of \(h^2\) we do it normally because we're deriving with respect to h. However, if we're deriving r, we can't assume r behaves exactly as h, so we have to add in the \(\Large \frac{\text{dr}}{\text{dh}}\) to account for that.

Answer 40

Gotcha, thanks. Okay so we can continue.

Answer 41

\(\Large \pi*r'*\sqrt{r^2+h^2} + \sqrt{r^2+h^2}'*\pi*r\)

Answer 42

Moving on to the right term:
\(\Large \pi*r'*\sqrt{r^2+h^2} + \sqrt{r^2+h^2}'*\pi*r\)
We'll use \(\sqrt{\text{ } }\) as our outside function and \((r^2+h^2) as our inside function.

Answer 43

Since \(\sqrt{x}\) is the same as \(x^{1/2}\), the derivative of the outside function comes out to be:
\(\Large \frac{1}{2}x^{-1/2}\) which is the same as \(\Large \frac{1}{2\sqrt{x}}\)

Answer 44

Derivative of the inside function is:
\(\Large (r^2+h^2)' = 2r*r' +2h\)

Answer 45

Following?

Answer 46

Yes i follow. Where does the r' come from again this time for the inside function? You took r^2 and made it 2r * r'? Will you always add that on, even when you still took the derivative of r in the term before?

Answer 47

Any time you're deriving r instead of h.

Answer 48

Gotcha. Okay

Answer 49

So we've settled the right term and we end up with:
\(\huge \pi*r'\sqrt{r^2+h^2} + \frac{(2r*r' +2h)*\pi*r}{2sqrt{r^2+h^2}}\)

Answer 50

The rest is algebra with the goal of solving for r'.

Answer 51

\(\huge \pi*r'\sqrt{r^2+h^2} + \frac{(2r*r' +2h)*\pi*r}{2\sqrt{r^2+h^2}}\)
(fixed latex)

Answer 52

Okay i follow you there. Still looks really messy for me, mind finishing the rest? sorry i really am bad at stuff like this.

Answer 53

We'll get a common denominator by multiplying the left term by \(\Large \frac{2\sqrt{r^2+h^2}}{2\sqrt{r^2+h^2}}\)

Answer 54

Oh god this is where it gets messy.

Answer 55

Actually it's where things start to clean up.
Giving us:
\(\huge \frac{2r'*(r^2+h^2)}{2\sqrt{r^2+h^2}} + \frac{(2r*r'+2h)*\pi*r}{2\sqrt{r^2+h^2}}\)

Answer 56

I think you forgot pi on the numerator for the left term?

Answer 57

\(\huge \frac{2\pi*r'*(r^2+h^2)}{2\sqrt{r^2+h^2}} + \frac{(2r*r'+2h)*\pi*r}{2\sqrt{r^2+h^2}}\)

Answer 58

Okay phew i follow still so far.

Answer 59

becomes 1 fraction
\(\huge \frac{2r'*(r^2+h^2) + (2r*r'+2h)*\pi*r}{2\sqrt{r^2+h^2}}\)

Answer 60

forgot pi again right?

Answer 61

\(\huge \frac{2\pi*r'*(r^2+h^2) + (2r*r'+2h)*\pi*r}{2\sqrt{r^2+h^2}}\)

Answer 62

Factor and cancel 2:
\(\huge \frac{\pi*r'*(r^2+h^2) + (r*r'+h)*\pi*r}{\sqrt{r^2+h^2}}\)

Answer 63

distribute:
\(\huge \frac{\pi*r^2*r'*+\pi*h^2*r' + \pi*r^2r'+\pi*h*r}{\sqrt{r^2+h^2}}\)

Answer 64

Group like terms:
\(\huge \frac{2\pi*r^2*r'*+\pi*h^2*r' +\pi*h*r}{\sqrt{r^2+h^2}}\)

Answer 65

Factor out r' from the first two terms:
\(\huge \frac{r'(2\pi*r^2+\pi*h^2) +\pi*h*r}{\sqrt{r^2+h^2}}\)

Answer 66

I can follow when you do it. But damn so much room for error. I guess i gotta make sure i do every step slowly.

Answer 67

Haha it takes a lot of practice. Take heart, I remember being a freshman in my calculus class and being utterly BEWILDERED at the pellet I was watching happen up on the board.

Answer 68

Heh yeah thats where i'm at. Thanks. Okay so now to simplify and solve for r'.

Answer 69

Right. The rest ain't so bad.

Answer 70

I hate to say it, but i really don't know how to solve for r'.

Answer 71

Multiply by that denominator on both sides. Subtract \(\pi*h*r\) next.

Answer 72

You're probabbly so annoyed.

Answer 73

Wait you can multiply it by 0 on the other side? I thought you could never do that, because then you just get rid of the denominator without changing anything.

Answer 74

Remember this was actually one half of an equation.
\(\huge \frac{r'(2\pi*r^2+\pi*h^2) +\pi*h*r}{\sqrt{r^2+h^2}} = 0\)

Answer 75

Yes i know, but i didn't think you can multiple the denominator to the other side (0) which cancles it out. Thought that was a rule you could never do.

Answer 76

Not really sure what you're thinking of, but you're allowed to multiply both sides of this equation by \(\sqrt{r^2+h^2}\)

Answer 77

Okay my bad.

Answer 78

Well the deno is gone, and -pihr is on the other side now.

Answer 79

I assume then just divide and isolate r'?

Answer 80

Good good. Yup. They factor and cancel pi.

Answer 81

Okay. Wow that was ridiculous.

Answer 82

Rest is easy just plugging in values. Thanks so much : )

Answer 83

My pleasure =)

Answer 84

Now the real question is if i can do that on my own.

Answer 85

Go back over the steps now while they're fresh in your head.

Answer 86

Already am thanks.

Answer 87

You'll be just fine. Are you in college? What are you majoring in?

Answer 88

Yeah sophmore about a month into calc 1. Computer engineering lol got a long way to go with math.