Question

Hi everyone! Can someone please tell me how they arrived at (dw/dx) = "stuff I don't understand"...I posted a pic from my book. Thanks! :o)

Answer 1

@ganeshie8 @hartnn

Answer 2

no matter how many times I try, I can't derive (1-n)y^-n(dy/dx)...I'm missing a step or something.

Answer 3

Hmm..it doesn't load.

Answer 4

Okay, nvm..

Answer 5

I just don;t get the (1-n) part coming down in front

Answer 6

Wait...I get it! geez...

Answer 7

omg...the "n" is just a place holder for a number...it's not even a variable...that's why they bring the whole thing down in front, it's no different than x^5 becoming 5x^4...omg

Answer 8

can anyone confirm my epiphany?

Answer 9

yes, its simple chain rule

1-n is just a constant

1-n-1 = -n is your exponent of y after the derivative

Answer 10

and chain rule because y is the function of x

do dy/dx would come out

Answer 11

Yay! @hartnn to the rescue :)

Answer 12

do you know how long I have spent trying to figure out how a problem with 3 variable magically went to 4? omg...grrr

Answer 13

LOL

Answer 14

thanks hartn! :o) here's your cookie! mmmgood

Answer 15

*eats the cookie*
nomnomnomnom

Answer 16

:o)

Answer 17

*burp*
oops sorry :P

Answer 18

no need to apologize at this hour...I can't see or hear anymore anyway!

Answer 19

thanks again!

Answer 20

do u see why n can't be 0 ?

Answer 21

because anything to the zero power is 1?

Answer 22

Anything (except 0) ^0 = 1

so n can't be 1,
that will make w = constant, dw/dx = 0

but why can't n be = 0
?

that will make w =y
whats the problem with that?

Answer 23

uhm...

Answer 24

is it because the presence of the zero will disrupt the subtle quantum fluctuations within the Higgs field and a singularity will form and grow expanding outward at the speed of light and mess up my hair?

Answer 25

you got it!
I am so proud of you :D

Answer 26

well I'm glad there are no zeros then....it takes forever to brush my hair! :o~

Answer 27

:P

Answer 28

if n was a zero then I'm guess it would violate the very rule of what makes a bernoulli DE a Bernoulli DE, y^n...if n is zero, then it would just be a DE in standard form?

Answer 29

i answered that as I am running away from the singularity! I better be right!

Answer 30

as always, thats correct :)
i just saw the section heading :P
Equations of Bernoulli, ...

Answer 31

I got it? really? I amaze myself at times! :o)

Answer 32

not me,
I expected you'll get it :)

Answer 33

thanks for the confidence...well back to the salt mines for me I guess..thanks Hartnn!