Question

u(y)=U[1−(y^2/b^2)]
find du/dy

Answer 1

\(u(y)=U[1-(\dfrac{y^2}{b^2})]\)

This is correct yes?
And you want to take the derivative of \(U[1-(\dfrac{y^2}{b^2})]\) with respect to \(y\) correct?

Answer 2

yes

Answer 3

Ok, let me see here.... whoa boy. I can see why you asked about this :)

Answer 4

i feel like i should know this but i have been out of school for quite some time

Answer 5

Now I am going to try this, but I provide no guarantee as to the accuracy of my techniques. I think we need to do implicit differentiation.

Answer 6

\(\dfrac{du}{dy}(U[1-(\dfrac{y^2}{b^2})])\)
\(=\dfrac{d}{dy}(U[1-(\dfrac{y^2}{b^2})])\)\)


\(\dfrac{d}{dy}(U[1-(\dfrac{y^2}{b^2})])=\dfrac{dU(u)}{du}\dfrac{du}{dy}\)
where, \(u=1-\dfrac{y^2}{b^2}\) and \(\dfrac{d}{dy}U(u)=U^{\prime}(u)\)

\(=(\dfrac{d}{dy}(1-\dfrac{y^2}{b^2}))U^{\prime}(1-\dfrac{y^2}{b^2})\)

Differentiate term by term and factor out any constants.

\([\dfrac{d}{dy}(1)-\dfrac{\dfrac{d}{dy}(y^2)}{b^2}]U^{\prime}(1-\dfrac{y^2}{b^2})\)

\(=-\dfrac{(\dfrac{d}{du}(y^2))U^{\prime}(1-\dfrac{y^2}{b^2})}{b^2}\)

\(=\color{blue}{-\dfrac{(2y)U^{\prime}(1-\dfrac{y^2}{b^2})}{b^2}}\)

\(\color{red}{\text{DISCLAIMER: Not entirely sure this is correct}}\)

Answer 7

@amistre64
Care to check my work?

Answer 8

@Luigi0210
Any thoughts?

Answer 9

Is U a function here??

Answer 10

@austinL

Answer 11

no, for this problem it is a consant

Answer 12

All I know is what was given to us. And so I just put \(U^{\prime}\).

Answer 13

@austinL: you made an easy question look difficult :P

Answer 14

that's my fault, I should have defined that better

Answer 15

yeah I understand that @mfred81

Answer 16

:)

Answer 17

@atlas
Was I correct though? That is the important thing!

Answer 18

Nope I don't think so

Answer 19

Well gosh darn it :P

Answer 20

k......did you assume U as a function of u??

Answer 21

I tried plugging it into my calculator to slve and it just spit it back out and said nope nope nope nope.
Wolfie agrees however, and also shows a different form.

Answer 22

solve*

Answer 23

u(y) = U[1-y^2/b^2]
I am assuming U is a function of U and b is a constant
Then
du/dy = U[-2y/b^2] + [1-y^2/b^2] U' (du/dy)

Answer 24

du/dy(1-U' + U'(y^2/b^2)) = -2U[y/b^2]

Answer 25

that's the answer assuming U is a direct function of U and dU/du = U'

Answer 26

thank you for your help guys! I am waaaay out of practice

Answer 27

so are we @mfred81