Question

find the second derivative of x^2-xy=1-y^2. express as a function of x and y only

Answer 1

what problem are u having?

Answer 2

i dont know how to do it

Answer 3

if u know how to differentiate,it is a pretty simple question then

Answer 4

i never did this with xy and 2 x

Answer 5

Derivative with respect to x or y?

Answer 6

yes can you show me a step by step of how to start it

Answer 7

That is a question for you..
Should we differentiate it with respect to x or y?

Answer 8

well it says express as a function of x and y , so i am not really sure

Answer 9

With respect to x then... Usual practice :|
e.g:
\[\frac{d}{dx}(xy)=y\frac{d}{dx}x + x\frac{d}{dx}y=...?\]

Answer 10

so should i brind all the numbers over to one side and do the first derivative then the second derivative

Answer 11

Better to do it step by step. Find the first derivative first.

Answer 12

Do you know implicit differentiation?

Answer 13

um i am bad with the names but probably not but do i wright the problem as 2x=-2x

Answer 14

because x^2 become 2x and -xy turn to one and 1 goes away and -y^2 truns into -2y

Answer 15

-xy won't turn to one and and -y^2 is not -2y

Answer 16

then can you show me the way the formula should look next so i can see what to do

Answer 17

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Answer 18

x^2-xy=1-y^2
Differentiate both sides with respect to x
\[\frac{d}{dx}(x^2-xy)=\frac{d}{dx}(1-y^2)\]\[\frac{d}{dx}(x^2)-\frac{d}{dx}(xy)=\frac{d}{dx}(1)-\frac{d}{dx}(y^2)\]\[2x-[y\frac{d}{dx}(x)+x\frac{d}{dx}(y)]=-2y\frac{dy}{dx}\]\[2x-y-x\frac{dy}{dx}=-2y\frac{dy}{dx}\]\[2x-y=-2y\frac{dy}{dx}+x\frac{dy}{dx}\]\[2x-y=(x-2y)\frac{dy}{dx}\]\[\frac{dy}{dx}=\frac{2x-y}{(x-2y)}\]That's what you'll get for the first derivative.

Answer 19

thank you that more clear i have to see it to learn it

Answer 20

so i have to now find the second derivative right

Answer 21

what do i do next?

Answer 22

You need to know implicit differentiation.

<Example 1> Make use product rule.
\[\frac{d}{dx}(xy) = x\frac{d}{dx}(y)+y\frac{d}{dx}(x)=x\frac{dy}{dx}+y\]
<Example 2> Make use of power rule and chain rule.
\[\frac{d}{dx}(y^n)=(n)y^{n-1}\frac{dy}{dx}\]

Yes, you have to differentiate the equation given twice.

Answer 23

of so for the 2 derivative do you get 3(x-y)/(x-2y)^2

Answer 24

No.. You need to differentiate it implicitly.

Answer 25

o... can u show me?

Answer 26

You need to read how I get the first derivative first.

Answer 27

If you don't understand, ask. Otherwise, you can't get the second derivative correct.

Answer 28

i thought that the way to get the first derivative is the same way to get the socond serivative its just done 2

Answer 29

Yes. but how do you get the first derivative?

Answer 30

dident you just show me you take the numver like 3^4 and do 4*3= 12^3 per number.

Answer 31

I showed you.. But you don't seem to get it.

Answer 32

WELCOME TO IMPLICIT DIFFERENTIATION 101 !
FEATURING TONIGHT'S GUEST @romio
So, Romio, it seems you're having trouble differentiating implicitly. Let's have recap. The following questions require you to differentiate with respect to x?

Ready?
Here we go.

What is...
\[\huge \frac{d}{dx}x^4\]

Answer 33

4x^3

Answer 34

The crowd is going WILD :D
Nicely done, @romio
Differentiating implicitly involves a twist... another variable, usually y, is involved
So, for the second question, what do you think is...
\[\huge \frac{d}{dx}y^4\]?

Remember, implicit differentiation assumes y is a function of x. So... your verdict, @romio ?

Answer 35

i have now (2x-y)/(x-2y) as the first derivative and apparently i tried the second derivative and i got that wrong

Answer 36

i need someone to show me a step by step for the second derivative

Answer 37

Well, if you must, and you're sure your first derivative is correct, just redo the steps, differentiating with respect to x assuming y is some function of x.
Whatever you did in getting the first derivative, it's the same as in getting the second.

Answer 38

i know i did it, but Callisto said my anwer was wrong so i dont know what i did wrong

Answer 39

Well, the only way to find out, is by redoing the question, again :D
Don't worry... we are under the ever-watchful eye of @Callisto ;)
Let's start with your equation...
\[\large x^2-xy=1-y^2\]
Now, your challenge, should you choose to accept it, is to differentiate both sides of the equation with respect to x. Are you up for it? :)
\[\huge \frac{d}{dx}(x^2-xy)=\frac{d}{dx}(1-y^2)\]

Answer 40

You'd do well to remember that differentiation distributes over addition and subtraction.
Good hunting :D

Answer 41

i got (2x-y)/(x-2y) for my first derivative

Answer 42

Yeah. But Callisto said it's wrong, didn't she? Well then, let's redo the question. I'm not saying I agree with her or disagree, but the only way to find out is to do it, step by step, as you said.

Don't be discouraged, just pause, do it slowly, and eventually, it'll be second nature. Now, about that question
\[\huge \frac{d}{dx}(x^2-xy)=\frac{d}{dx}(1-y^2)\]

Answer 43

no the second derivative the first was wright

Answer 44

Was it now?
Can you tell me what
\[\Large \frac{d}{dx}y^4\]is?

Answer 45

4y^3

Answer 46

please can you show me to to turn my first derivative into a second derivative i did it like 20 times and i am doing something wrong

Answer 47

I see... what the problem is :)
Indeed
\[\Large 4y^3=\frac{d}{dy}y^4 \]but
\[\Large \frac{d}{dx}y^4=?\]

Answer 48

The thing is, we are differentiating along x, and y is a FUNCTION of x, so a good thing to remember, is to differentiate y AS YOU WOULD DO if it were x, however, when you're finished, you append dy/dx (or y') to it, due to the chain rule...

\[\Large \frac{d}{dx}y^4=4y^3\frac{dy}{dx}=4y^3y'\]
I prefer the former, the one involving dy/dx, it's more... in-your-face :D

Answer 49

ok so i took my first derivative and i did ((x-2y)(2-1)-[(2x-y)(1-2)])/(x-2y)^2 is this right my first derivatinv was (2x-y)/(x-2y)

Answer 50

then i worked it out but am i on the right track

Answer 51

Well, it seems like you've gotten it.

Answer 52

yea i know but my answer was wrong can you do it and tell me what you got because i must be doing some kind of multiplication error or something like as brakedown from where i showed you

Answer 53

Well, your first derivative is correct... how did you go about getting your second derivative?

Answer 54

i used this formula ((x-2y)(2-1)-[(2x-y)(1-2)])/(x-2y)^2

Answer 55

what is the answer T_T

Answer 56

The formula you used is not correct.

Answer 57

o? what, why not i thought it was?.... ok well can you tell me what the correct formula to use for this is then?

Answer 58

What formula? You just differentiate, don't you?

Answer 59

I've explained how to do it twice. Please refer to the above comments.

Answer 60

so is the answer -3y/(x-2y)^2 for the second derivative and if that is the answer is that all i have to do

Answer 61

Also, quotient rule...
\[\Large \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\]

Answer 62

it says express a function of x and y only

Answer 63

@romio
Wolfram Alpha?
Two things:
One, you can't do that on exams...
Two, that gives you what's called the PARTIAL derivative of (2x-y)/(x-2y)
\[\huge \frac{\partial}{\partial x}\frac{2x-y}{x-2y}=\frac{-3y}{(x-2y)^2}\]

Answer 64

so is my answer wrong

Answer 65

Yep... as far as the second derivative goes. Look, you can't escape learning how to differentiate implicitly, so just face the music ;)

Answer 66

ok so can you tell me what the answer is so i can figure out what i am doing wrong

Answer 67

Very poor choice of words when you know a Moderator is watching :D
The Code of Conduct encourages you to arrive at the answer yourself, with, of course, some help from the users :)

Answer 68

Hmm... Do you know where you get it wrong in the first derivative, after I have given you the answer?

Answer 69

partly, it will also be a good way to check if i did it wright because i dont know the answer so i wont be able to know if i am doing it right or wrong.

Answer 70

Yeah, but there's a better way... actually doing it.... you know, taking
\[\huge \frac{dy}{dx}=\frac{2x-y}{x-2y}\]and differtiating both sides? :D
C'mon, I know you can do it ;)

Answer 71

@romio has left?