Question

Good evening OS! Now let's do some Calc...
I need assistance finding the derivative of this equation:
"C = [40,000(sqrt.(144 + y^2))] + [30,000(20 - y^2)]"

Any and all help is greatly appreciated! :)

Answer 1

\[C=40000\sqrt{144+y^2}+30000(2-y^2)\]

Answer 2

try product rule and chain rule

Answer 3

Mainly, my question is, Can I distribute the 40,000? and Do I distribute the square root to the 144 & y^2, before taking the derivative?

Answer 4

how would you re-write it so it will be manageable using the rules?

Answer 5

Yep, that's the equation, prettier :)
I know how to take derivatives, but with all of the possible algebraic action possible in that first term, I'm going from the aspect of what's RIGHT, step-wise from a theory-based approach.

Answer 6

Yeah, I know, square root = to the half power

Answer 7

absolutely you can use chain rule in the first term

Answer 8

So just take the derivative right away? Is 40,000 considered a constant? It just gets multiplied by 1/2 right?

Answer 9

yeah of course
it is why I asked if you want to rewrite it so you can see it clearly

Answer 10

I mean, I would even rewrite the whole thing so it would be manageable
4000 and 3000? can it be reduced?

Answer 11

40000 and 30000

Answer 12

Alright, sweet! :) I was just unsure of, technically, what should get done first, or what was free to be simplified before taking a derivative. Thank you!

Answer 13

They are specific amounts ($) crucial to the (word) problem. :/

Answer 14

May the second half of that equation be simplified before finding C' ? The 30,000 being distributed that is.

Answer 15

no matter the magnitude, the derivative of a constant is?

Answer 16

0, but can I write it as "(blah, blah) 600,000 - 30,000y" = "(blah, blah) - 30,000y + 600,000"? And then yeah, I see the 600,000 would become 0 in C'. :)

Answer 17

Oh, sorry, and a + after the 1st "(blah, blah)

Answer 18

yes

Answer 19

Sweet, thanks !

Answer 20

It's like OS is more of a practical teacher to me than my classroom teacher at school, lol. You sure help more, and faster ;)

Answer 21

Haha, come to think of it, I recall that I've pulled up OS on my computer at school during class when the teacher was flaking out, and gotten help on stuff that way. Ironic...

Answer 22

professors give you the more detailed work, we here just simply provide the pieces in the gaps in learning

Answer 23

Haha, true... But at least I can ask a question, and get an answer. My prof is... an unresponsive recipient ;)

Answer 24

Thanks again!

Answer 25

I know I closed the question, but now that I've solved the first derivative for y and all, I find myself perplexed by the answer. Would you mind working this though, that I may compare my work /answer? @nincompoop :)

Answer 26

@satellite73 would you be willing to help me out?