Question

Higher order derivative question...

Answer 1

\[y^2=x^3\]find\[\frac{ d^2y }{ dx^2 }\]

Answer 2

I get \[\frac{ dy }{ dx }=\frac{ 3x^2 }{ 2y }\]

Answer 3

When taking the second derivative I'm not getting what they're getting.

Answer 4

I can walk you through what I'm doing and you can see where I go wrong?

Answer 5

wait woops

Answer 6

mistaken the y for an x

Answer 7

When taking the next higher order derivative I get\[\frac{ d }{ dx }\frac{dy}{dx}=\frac{ 2y*\frac{ d }{ dx }[3x^2]-3x^2*\frac{ d }{ dx }[2y] }{ 4y^2 }\]

Answer 8

Which then leads me to\[\frac{ d^2y }{ dx^2 }=\frac{2y*6x-3x^2*2\frac{ dy }{ dx }}{4y^2}\]

Answer 9

That simplifies to\[\frac{ d^2y }{ dx^2 }=\frac{12xy-6x^2*\frac{dy}{dx}}{4y^2}\]

Answer 10

dude, what on earth are you doing

Answer 11

Then I can substitute for dy/dx

Answer 12

LOL... wish I knew!

Answer 13

I think he used the rule for add/sub for derivatives

Answer 14

I used the quotient rule

Answer 15

Here is how the instructor showed the work. I understand everything up to the point to where they get 4x^3 in the denominator when taking the second derivative.

Answer 16

So taking the first implicit derivative is easy... taking the second should be easy but I just don't see that 4x^3 in the denominator. That's what is throwing me. I think my brain is fried.

Answer 17

oh, I read y^2 = x^2 and second derivative of y

Answer 18

yeah, you're doing it right

Answer 19

I see that I'm doing it right up to this point. However, this is where my work starts to differ from the attachment.

Answer 20

y^2 = x^3
So 4y^2 in the denominator is substituted with 4x^3

Answer 21

that made so much sense if you look at the original equation you see what ranga said

Answer 22

HOLY CRAP.

Answer 23

I've spent an hour looking at this thing and I totally missed that.

Answer 24

Thank you!

Answer 25

You are welcome.

Answer 26

Why did they choose to do that? Do you have to do that?

Answer 27

I would solve this problem much easier than they did.
y^2 = x^3. Find the third derivative y'''.
y^2 = x^3
y = x^(3/2)
y' = 3/2x^(1/2)
y'' = 3/2 * 1/2 * x^(-1/2) = 3/4 * x^(-1/2)
y''' = 3/4 * (-1/2) * x^(-3/2) = -3/8 * x^(-3/2)
= -3/8 * 1 / x^(3/2) = -3/8 * 1/ { (x^3)^(1/2) = -3/8 * 1 / (y^2)^(1/2)
= -3/8y

Answer 28

They did that to simplify. It is better to have as few y on the right as possible because each time you take the derivative it will be dy/dx and you will have to substitute for that. So convert y to x whenever you get a chance. Above I chose to do it in the first step. They chose to do much later.

Answer 29

That makes sense! Thanks again!

Answer 30

I've since worked the problem a few times a few different ways just to get a better understanding. All total I've spent a couple of hours just on this one problem. I couldn't let it go! :) Thanks!