Question

If \[\large\rm x=e^t\ \]
Show that (Done) \[\large\rm x\frac{dy}{dx}=\frac{dy}{dt} \ \]
And (Need help on this)
\[\large\rm x\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt} \]

And please also explain if we are going to describe \[\large\rm \frac{d^2y}{dx^2},\frac{dy^2}{dx^2},d^2y \] in English, what would it be? (Like change in y with respect to.....)

Answer 1

@zepdrix

Answer 2

Is it not that
\(\dfrac{d}{dt}(\dfrac{dy}{dt})=\dfrac{dy}{dx}+x\dfrac{d^2y}{dx^2}\)
and you got the answer?

Answer 3

How do you got that. Can you explain?

Answer 4

\(\dfrac{d^2y}{dx^2}\) is just second derivative of y with respect to x.

Answer 5

The left hand side is just a factor of x and dy/dx. Take product rule.

Answer 6

@zepdrix now, your turn. hehehe...

Answer 7

x=e^t

where is y...?

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Loser66
\(\dfrac{d^2y}{dx^2}\) is just second derivative of y with respect to x.
\(\color{#0cbb34}{\text{End of Quote}}\)
Yeah but what does it mean in English? Like dy/dx is said to be change in y with respect to change in x

Answer 9

@zepdrix Please, explain him.

Answer 10

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Loser66
Is it not that
\(\dfrac{d}{dt}(\dfrac{dy}{dt})=\dfrac{dy}{dx}+x\dfrac{d^2y}{dx^2}\)
and you got the answer?
\(\color{#0cbb34}{\text{End of Quote}}\)
It is acquired by doing right?
\[\large\rm \frac{d}{dx} (x\frac{dy}{dx})=\frac{d}{dt} (\frac{dy}{dt} ) \]

Answer 11

yup

Answer 12

about second derivative!! I think you should understand what it means!!

Answer 13

Suppose you have a function y =x^2, then dy / dx is the tangent of the curve, right?

Answer 14

derivative measures change in y with respect to x, so,
second derivative measures change in (change in y with respect to x).

It's measuring the change in the change.
It makes a little more sense if you think of it in terms of movement I suppose.

Like if you're measuring the movement of a particle over time...
derivative measures the moment over time (we call this velocity),
second derivative measures the change in velocity (we call this acceleration).

Answer 15

Yes I am aware that it is a derivative of dy/dx with respect to x. But how should I explain in Enlglish terms

Answer 16

@zepdrix thank you ;)

Answer 17

derivative measures the movement* over time (we call this velocity),

blah typo

Answer 18

\(\color{#0cbb34}{\text{Originally Posted by}}\) @zepdrix
derivative measures change in y with respect to x, so,
second derivative measures change in (change in y with respect to x).

It's measuring the change in the change.
It makes a little more sense if you think of it in terms of movement I suppose.

Like if you're measuring the movement of a particle over time...
derivative measures the moment over time (we call this velocity),
second derivative measures the change in velocity (we call this acceleration).
\(\color{#0cbb34}{\text{End of Quote}}\)
Yes that's exactly what I was thinking change in change in y with respect to x. But what about \[ \frac{dy^2}{dx^2 } \]

Answer 19

Change in change in y divided by change in change in x, perhaps?

Answer 20

You're asking what the difference is between \(\large\rm \frac{d^2y}{dx^2}\) and \(\large\rm \frac{dy^2}{dx^2}\) ?

Answer 21

Yes

Answer 22

This \(\large\rm \frac{dy^2}{dx^2}\) actually doesn't mean anything.
It's not properly written.

\(\large\rm \frac{d^2y}{dx^2}\) is the second derivative of y.

\(\large\rm \frac{d^2y^2}{dx^2}\) is the second derivative of y^2.

but if your top d doesn't have a second power,
then it's just... not a thing D:

It can't be second derivative in the bottom,
and first derivative notation in the numerator.

At least that's my understanding of it :d

Answer 23

I guess you could do something like this,\[\large\rm \frac{dy^2}{dx^2}=\frac{dy^2}{dx}\cdot\frac{1}{dx}\]So it's the derivative of y^2 with respect to x,
divided by the instantaneous rate of change of x.

I don't know if that has any real meaning though.. hmm

Answer 24

\[\large\rm x^2=y^2\ \]
\[\large\rm 2x ~ dx=2y~ dy\ \]
\[\large\rm 2 ~ dx^2=2~ dy^2\ \]
\[\large\rm 2 =2~ \frac{dy^2}{dx^2}
\ \]

Answer 25

So according to you that dy^2/dx^2 doesn't mean anything what does it mean here?

Answer 26

And one last thing
Is this true? And if not what does it equal to?
\[\large\rm (\frac{dy}{dx})^2 =\frac{d^2y^2}{d^2x^2} \]

Answer 27

\(x^2=y^2\\x =\pm y\)

Answer 28

Wow people do math this early ._.

Answer 29

Boy these are good questions :D
I'm a little stumped lol.
Still thinking...

Answer 30

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Loser66
\(x^2=y^2\\x =\pm y\)
\(\color{#0cbb34}{\text{End of Quote}}\)
I didn't take a root, I differentiated them

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Sainath
Wow people do math this early ._.
\(\color{#0cbb34}{\text{End of Quote}}\)
Its 6 pm in my region

\(\color{#0cbb34}{\text{Originally Posted by}}\) @zepdrix
Boy these are good questions
I'm a little stumped lol.
Still thinking...
\(\color{#0cbb34}{\text{End of Quote}}\)
And I was thinking that you guys aren't replying because you consider my question s stupid. Thank you for clarification.

Answer 31

First off, your derivative is nonsense to me.
if \(x^2=y^2\)
\(\dfrac{d}{dx}(x^2)=\dfrac{d}{dy}(y^2)\)
That gives us 2x = 2y
It doesn't give 2xdx =2ydy

Answer 32

if you take second derivative, again, you get 2=2 , period

Answer 33

You don't remember doing a little bit of that back in calculus Lose? :o
Finding rate of change in terms of differentials.\[\large\rm y=3x+2\qquad\to\qquad dy=3dx\]

Answer 34

Oh apologies. Here is a modified version
\[\large\rm x=y\\]
\[\large\rm d/dx x=d/dyy\\]
\[\large\rm d/dx dx=d/dy dy\\]
\[\large\rm dx^2= dy^2\\]
\[\large\rm 1 = \frac{dy^2}{dx^2} \\]

Answer 35

@Loser66 Your last comments was for the previous version right?

Answer 36

I like your previous example better actually.
\(\large\rm x^2=y^2\)

Let's do it the normal way first.
Differentiating each side `implicitly` with respect to x,
\(\large\rm 2x=2yy'\)
Solving for our derivative gives us,\[\large\rm y'=\frac{x}{y}\]squaring each side gives us,\[\large\rm (y')^2=\frac{x^2}{y^2}\]

Let's check out what happens when we do it the other way,\[\large\rm 2xdx=2ydy\]Dividing each side by 2,
squaring each side,\[\large\rm x^2dx^2=y^2dy^2\]Solving for ... something...\[\large\rm \frac{x^2}{y^2}=\frac{dy^2}{dx^2}\]So `in this particular case`,
we determined that,\[\large\rm \frac{dy^2}{dx^2}=(y')^2=\left(\frac{dy}{dx}\right)^2\]

Answer 37

So yesss it actually does have some meaning, ok my mistake.
It's the square of the differentials...
so when you divide them, it's the square of the derivative!

Answer 38

Which does not necessarily have anything to do with the second derivative.

Answer 39

@zepdrix Thank you very much for explaining

Answer 40

\[\large\rm \frac{dy^2}{dx^2}=\left[\frac{dy}{dx}\right]^2=\left[f'(x)\right]^2\]

Answer 41

Which is not the same as this\[\large\rm \frac{d^2y}{dx^2}\]except in some rare cases.

Answer 42

To me, if y^2 =x^2
then dy^2/dx^2 =1

Answer 43

and d^2y/dx^2 =0

Answer 44

Lose, remember doing differential math in integral calc with your u-subs? :)\[\large\rm u=x^2+3\qquad\to\qquad du=2x~dx\]

Answer 45

I'm not sure of the exact meaning :P
I just remember doing that sort of thing lol

Answer 46

let me explain!!

Answer 47

if y^2 =x^2 if you take derivative of y w.r.t x, then you get what you said above
but if you take derivative of (y^2) with respect to the whole (x^2) then it is =1

Answer 48

after then, the second derivative of (y^2) w.r.t (x^2) =0 because it is derivative of a constant.

Answer 49

@zepdrix In your example, it sounds like this
\[\large\rm u = x^2+3\ \]
\[\large\rm \frac{d}{dx}u=\frac{d}{dx} (x^2+3)\ \]
\[\large\rm \frac{du}{dx}=2x\ \]
\[\large\rm du=2x dx\ \]

Answer 50

@zepdrix Thank you very much for explaining the meaning of \(\large\rm \frac{dy^2}{dx^2} \) and also how to incorporate LaTex so the terms come in line with text

Answer 51

Ya this was a fun question :)
I need to brush up on some Calc stuff