What is the derivative for x^2 + y^2 + ln(2) = xy?

[antinerd]

Homework Statement

Find dy/dx in terms of x and y

Homework Equations

x^2 + y^2 + ln(2) = xy

The Attempt at a Solution

I'm pretty sure I did it right, but I am wondering if I was supposed to take the derivative of everything:

2x + 2y = 1

Is that the correct derivative? If so, then I'm good to go...

 

[rocomath]

ln(2) and xy can be derived as well

 

[antinerd]

ln(2) and xy can be derived as well

Didn't I already derive xy? It's 1 right?

And isn't ln(2) a constant?

 

[rocomath]

xy = product rule ... are you doing implicit/explicit differentiation? y' if you are.

ln(2) ... i forgot, been a while since I've taken Calculus

 

[antinerd]

xy = product rule ... are you doing implicit/explicit differentiation? y' if you are.

ln(2) ... i forgot, been a while since I've taken Calculus

It just says: "Find dy/dx"

So, OK, let me correct the dy/dx xy:

so it's y + x

Now, can the answer be written thus:

x + y = 1 ??

And would I have to write it like this:

dy/dx x+y = 1 dy/dx

 

[d_leet]

It just says: "Find dy/dx"

So, OK, let me correct the dy/dx xy:

so it's y + x

Now, can the answer be written thus:

x + y = 1 ??

And would I have to write it like this:

dy/dx x+y = 1 dy/dx

Where is this "1" coming from? And that still isn't the correct derivative of xy, you need to use the product rule. What is the derivative of y with respect to x?

 

[rocomath]

have you learned the product rule?

f(x)g(x) = f'(x)g(x) + g'(x)f(x)

 

[abelian jeff]

You need to take partial derivatives, and in the process of doing this, you will get dy/dx 's in your equation. Then group like terms and solve for dy/dx. Have you learned about partial derivatives? The product rule mention above will be helpful in the process, as well.

 

[antinerd]

You need to take partial derivatives, and in the process of doing this, you will get dy/dx 's in your equation. Then group like terms and solve for dy/dx. Have you learned about partial derivatives? The product rule mention above will be helpful in the process, as well.

Alright hold up.

 

[d_leet]

You need to take partial derivatives, and in the process of doing this, you will get dy/dx 's in your equation. Then group like terms and solve for dy/dx. Have you learned about partial derivatives? The product rule mention above will be helpful in the process, as well.

I'd imagine this problm is showing up in the context of a calulus one class after a lesson on implicit differentiation, hence the original poster has probably never encountered partial derivatives before, and has no reason at all to find them for this problem, I don't understand why you think he needs to take partial derivatives.

 

[abelian jeff]

Crap, you're right, I made a typo. I meant to type implicit, not partial. Apologies.

 

[Saladsamurai]

have you learned the product rule?

f(x)g(x) = f'(x)g(x) + g'(x)f(x)

Not to be nitpicky, but I think what rocophysics meant was
(fg)'(x)=f'(x)g(x) + g'(x)f(x)

I don't go looking to correct peoples' mistakes, however I know how difficult learning new notations can be. If we all start out on the same page, it will be easier down the road.

Casey

 

[antinerd]

The thing that I'm not getting is how to get from

2x + 2y + ln2 = xy

to

d/dx (2x + 2y + ln2) = xy

and what to do after that, and what to do with the other side of the equation...

 

[antinerd]

Can someone show me a small example of how to do these derivatives?

 

[abelian jeff]

If you take the derivative with respect to x of the left hand side of the equation, you have to take it of the right hand side, too.

However, simply writing d/dx on the left side there isn't enough. You need to take the derivative with respect to x to each of the summands on the left side (and, of course, the right side of the equation too).

For instance, the derivative w.r.t. x of $$5x^3$$ would be $$15x^2(\frac{dx}{dx})=15x^2.$$
The derivative w.r.t. x of $$4y^3$$, however would be $$12y^2(\frac{dx}{dy})$$. The $$\frac{dx}{dy}$$ remains in this case.

You have learned about implicit differentiation, correct?

 

[Saladsamurai]

The thing that I'm not getting is how to get from

2x + 2y + ln2 = xy

to

d/dx (2x + 2y + ln2) = xy

and what to do after that, and what to do with the other side of the equation...

Let's start over. You are being asked to implicitly differentiate the function
$$x^2+y^2+ln(2)=xy$$

Now how about an easier one to start. Can you try to implicitly differentiate
$$x^2+y^2=1$$ with respect to x?

 

[Dick]

You want d/dx(x^2 + y^2 + ln(2)) = d/dx(xy). Split this into four derivatives and I'm sure anyone here will jump on anyone you get wrong.

 

[antinerd]

If you take the derivative with respect to x of the left hand side of the equation, you have to take it of the right hand side, too.

However, simply writing d/dx on the left side there isn't enough. You need to take the derivative with respect to x to each of the summands on the left side (and, of course, the right side of the equation too).

For instance, the derivative w.r.t. x of 5x^3 would be 15x^2(dx/dx)=15x^2.
The derivative w.r.t. x of 4y^3, however would be 12y^2(dy/dx). The dy/dx remains in this case.

You have learned about implicit differentiation, correct?

Yeah, I have. But I suck at life.

So what about this for starters:

d/dx (x^2 + y^2 + ln2) = (xy) d/dx

would I then do:

2 + 2y (d/dx) + 0 = xy d/dx

?

 

[antinerd]

Let's start over. You are being asked to implicitly differentiate the function
$$x^2+y^2+ln(2)=xy$$

Now how about an easier one to start. Can you try to implicitly differentiate
$$x^2+y^2=1$$ with respect to x?

$$x^2+y^2=1$$

2x + y^2 = 1

?

 

[antinerd]

And what about finding dy/dx in terms of x AND y?

 

[Saladsamurai]

$$x^2+y^2=1$$

2x + y^2 = 1

?

2x is good.
Now what about $$y^2$$? what happened to its derivative? Apply the power rule $$nx^{n-1}$$ and then we'll apply the chain rule.

Also what is the derivative of a constant (1)?
Remember that when differentiating implicitly, we are differentiaiting the ENTIRE function with respect to x.

 

[rocomath]

Not to be nitpicky, but I think what rocophysics meant was
(fg)'(x)=f'(x)g(x) + g'(x)f(x)

I don't go looking to correct peoples' mistakes, however I know how difficult learning new notations can be. If we all start out on the same page, it will be easier down the road.

Casey

no prob, I'm picky as well

 

[antinerd]

2x is good.
Now what about $$y^2$$? what happened to its derivative? Apply the power rule $$nx^{n-1}$$ and then we'll apply the chain rule.

Also what is the derivative of a constant (1)?
Remember that when differentiating implicitly, we are differentiaiting the ENTIRE function with respect to x.

But if it's with RESPECT to x, don't you just derive the terms with x?

But to answer your question, it's:

2x + 2y = 0

correct?

 

[d_leet]

But if it's with RESPECT to x, don't you just derive the terms with x?

But to answer your question, it's:

2x + 2y = 0

correct?

No that is not true, you're forgetting the chain rule. What is the derivative of y² if y is a function of x?

 

[learningphysics]

Remember that d/dx (y) = dy/dx

To take the derivative of y^3 for example you use the chain rule...
$$\frac{d}{dx}(y^3) = \frac{d}{dy}(y^3) * \frac{dy}{dx} = 3y^2 * \frac{dy}{dx}$$


Example of product rule...
$$\frac{d}{dx}(x^3y^2) = <br /> (\frac{d}{dx}(x^3))(y^2) + (x^3)(\frac{d}{dx}(y^2))<br /> =3x^2y^2 + x^3(2y*\frac{dy}{dx})<br /> =3x^2y^2 + 2x^3y*\frac{dy}{dx}$$

 

[Saladsamurai]

But if it's with RESPECT to x, don't you just derive the terms with x?

correct?

No because everything except the constant varies with x. if you wanted you could rewrite the equation EXPLICITLY in terms of y to get y=sqrt(1-x^2) and then take the derivative...but we want to do it IMPLICITLY.

WRT x means "as x changes), not just the terms involving x 'matter'

But if it's with RESPECT to x, don't you just derive the terms with x?

But to answer your question, it's:

2x + 2y = 0

correct?

You are close, but we want the derivative of y^2 wrt x, so we need to apply the chain rule. Do you know the chain rule d/dx [u^2]=2u*du/dx

 

[Saladsamurai]

So in the case of y^2 it is 2y times the derivative of y wrt x ...and that is just dy/dx

 

[bomba923]

Homework Equations

x^2 + y^2 + ln(2) = xy

Are you sure you wrote that down correctly?

Because $$x^2 + y^2 + \ln 2 = xy$$ has no real solutions...

*Proof: Assume $$x,y \in \mathbb{R}$$
then $$x^2 + y^2 + \ln 2 = xy \Rightarrow x^2 + y^2 < xy$$

>However, since $x=y \vee x \ne y$,
$$\begin{gathered}<br /> {\text{1)}}\;x = y \Rightarrow x^2 + y^2 < xy \Leftrightarrow 2x^2 < x^2 ,\;{\text{which}}\;{\text{is a contradiction}} \hfill \\<br /> {\text{2)}}\;x \ne y \Rightarrow \left| x \right| < \left| y \right| \vee \left| x \right| > \left| y \right| \hfill \\<br /> {\text{ }}\left| x \right| < \left| y \right| \Rightarrow xy < y^2 \Rightarrow xy < x^2 + y^2 \hfill \\<br /> {\text{ }}\left| x \right| > \left| y \right| \Rightarrow x^2 > xy \Rightarrow x^2 + y^2 > xy \hfill \\<br /> {\text{ Both of these contradict }}x^2 + y^2 < xy \hfill \\ <br /> \end{gathered}$$

 

[learningphysics]

Are you sure you wrote that down correctly?

Because $$x^2 + y^2 + \ln 2 = xy$$ has no real solutions...

*Proof: Assume $$x,y \in \mathbb{R}$$
then $$x^2 + y^2 + \ln 2 = xy \Rightarrow 0 \leqslant x^2 + y^2 < xy \Rightarrow 0 < xy$$

>However, since $x=y \vee x \ne y$,
$$\begin{gathered}<br /> {\text{1)}}\;x = y \Rightarrow x^2 + y^2 < xy \Leftrightarrow 2x^2 < x^2 ,\;{\text{which}}\;{\text{is a contradiction}} \hfill \\<br /> {\text{2)}}\;x \ne y \Rightarrow \left| x \right| < \left| y \right| \vee \left| x \right| > \left| y \right| \hfill \\<br /> {\text{ }}\left| x \right| < \left| y \right| \Rightarrow \left| {xy} \right| < y^2 \Rightarrow \left| {xy} \right| < x^2 + y^2 \Rightarrow xy < x^2 + y^2 \because xy > 0 \hfill \\<br /> {\text{ }}\left| x \right| > \left| y \right| \Rightarrow x^2 > \left| {xy} \right| \Rightarrow x^2 + y^2 > \left| {xy} \right| \Rightarrow x^2 + y^2 > xy\because xy > 0 \hfill \\<br /> {\text{ Both of these contradict }}x^2 + y^2 < xy \hfill \\ <br /> \end{gathered}$$

Good point! I guess we can assume x and y are complex... I don't think it affects the implicit differentiation does it?

 

[antinerd]

Am I the only one who found this strange?

Are you sure you wrote that down correctly?

Because $$x^2 + y^2 + \ln 2 = xy$$ has no real solutions...

*Proof: Assume $$x,y \in \mathbb{R}$$
then $$x^2 + y^2 + \ln 2 = xy \Rightarrow x^2 + y^2 < xy$$

>However, since $x=y \vee x \ne y$,
$$\begin{gathered}<br /> {\text{1)}}\;x = y \Rightarrow x^2 + y^2 < xy \Leftrightarrow 2x^2 < x^2 ,\;{\text{which}}\;{\text{is a contradiction}} \hfill \\<br /> {\text{2)}}\;x \ne y \Rightarrow \left| x \right| < \left| y \right| \vee \left| x \right| > \left| y \right| \hfill \\<br /> {\text{ }}\left| x \right| < \left| y \right| \Rightarrow xy < y^2 \Rightarrow xy < x^2 + y^2 \hfill \\<br /> {\text{ }}\left| x \right| > \left| y \right| \Rightarrow x^2 > xy \Rightarrow x^2 + y^2 > xy \hfill \\<br /> {\text{ Both of these contradict }}x^2 + y^2 < xy \hfill \\ <br /> \end{gathered}$$

Yeah, well, that's the question that's written down...

Hmm... I guess it doesn't really matter. Does it even affect what I will get as a derivative?