Question

How to find the 2nd derivative of 2x^2-3y^2-4?

Answer 1

You find the first derivative the differentiate the first derivative.

Answer 2

Ah well is it 4x-6y dy/dx?

Answer 3

That's the first derivative. Now differentiate it again ^^

Answer 4

err I don't know how. Lemme try that again. Can you please explain how -3y^2 disappeared in this solution? This was the solution from a second derivative calculator

Answer 5

It's the same equation. I didn't understand why -3y^2 was treated as a constant

Answer 6

Because the calculator is dumb.

Answer 7

I think you are not dealing with partial derivatives right?

Answer 8

Me neither. It would have made sense if it was \[\Large \frac{\partial}{\partial x}\] though.

I agree with Thomas, your calculator is dumb. Trust your human abilities :P

Answer 9

Okay, do you know the product rule? ^^

Answer 10

fg'+gf'?

Answer 11

One thing to know it in theory, and another to practice it :P
Let's do that right now.

Time to differentiate
\[\Large 4x - 6y \frac{dy}{dx}\]

So... ready?

Answer 12

ok, I'm ready

Answer 13

Good. Time to differentiate.
\[\Large \frac{d}{dx}\left[4x - 6y \frac{dy}{dx}\right]\]
Let's start with the 'easy' bit. Differentiate 4x, you get?

Answer 14

4

Answer 15

Good :P
\[\Large 4 - \frac{d}{dx}\left[6y\frac{dy}{dx}\right]\]
Now you have the derivative of a product, 6y and dy/dx

Differentiate.

Answer 16

Hey, are you having trouble?
Put down your best try, don't be afraid to make mistakes. OR to tell me that you're stuck.

It's MUCH better that you get stuck/commit errors here than on an exam. ^_^

Answer 17

Well first I took out the 6 and have y dy/dx left to derive

Answer 18

then applying the product rule on y dy/dx I got stuck

Answer 19

\[4-6(\frac{d}{dx}(y\frac{dy}{dx}))\]

Answer 20

\[
\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^2y}{dx^2}
\]
Don't think too much.

Answer 21

(fg)' = f'g + fg'
It's all a matter of knowing exactly what your f and g are.

Here's an idea. Your "f" here is y, and your "g" here is dy/dx

Answer 22

\[y (\frac{d}{dx}\frac{dy}{dx})+\frac{dy}{dx}(\frac{d}{dx}y)\] like this?

Answer 23

Yup. Pretty much.

Answer 24

Be careful :)

Answer 25

how am I supposed to do \[\frac{d}{dx}y\]

Answer 26

Uhh... what indeed :P
\[\Large \frac{d}{dx}y \qquad \rightarrow \qquad \frac{dy}{dx}\]
don't you think? :P

Answer 27

hmm so there's no special meaning behind it =____=

Answer 28

nope. nothing more you can do about that.

Answer 29

then I have \[4-6(\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2)\]

Answer 30

There you go ^^
That wasn't so hard, was it? :P

Answer 31

turns out we were wrong ._.

Answer 32

I failed the homework and showed me the solution

Answer 33

We weren't. You just didn't show the entire problem :P
That = 4 matters.

Answer 34

Had we started with
\[\Large 2x^2 - 3y^2 = 4\]
We would have arrived there too ^^

Answer 35

Oh well

Answer 36

Thanks for the help :)

Answer 37

Sorry we didn't arrive at your desired solution. Next time (if ever) show a screenshot of your problem ^^