1
$\begingroup$

One of my classmates said that for $x^2+y^2=1$, to find $\frac{dy}{dx}$, the following method can be used:

First rearrange the equation, $$x^2+y^2-1=0$$. Then assume $$w=x^2+y^2-1$$ $$\frac{dy}{dx}=\frac{dw}{dx} \div \frac{dw}{dy}$$

Also, when finding $\frac{dw}{dx}$, $y$ is considered a constant. And similarly, when finding $\frac{dw}{dy}$, $x$ is considered a constant.

Let me proceed. $\frac{dw}{dx}=2x$ regarding $y$ as a constant. $\frac{dw}{dy}=2y$ regarding $x$ as a constant. $\frac{dy}{dx}=2x/2y$ which is the negative of the correct answer.

It seems like it applies for all the equations.

I think his methods is completely unreasonable but by using his method the result is always the negative of the correct result. Why????

$\endgroup$
  • $\begingroup$ No, when you find $\frac{dw}{dw}$ you DO NOT treat $y$ as a constant, but as a function of $x$, which you are differentiating with respect to. You're thinking of a partial derivative. $\endgroup$ – Andrew Li Feb 27 '18 at 16:16
2
$\begingroup$

What he is showing you is the result of partial differentiation. In partial differentiation, for a function $$u = f(x, y),$$ then $$du = \partial_x u + \partial_y u$$ where $\partial_x u$ is the partial differential of $u$ with respect to $x$ (i.e., differentiating with all variables but $x$ held constant).

Since, in your case, $u$ (and therefore $du$) is zero, we can then say:

$$ 0 = \partial_x u + \partial_y u $$

Now, if we solve for one of them, we get:

$$ \partial_x u = -\partial_y u $$

Or, in other words,

$$ \frac{\partial_x u}{\partial_y u} = -1 $$

That's for differentials. The partial derivative of $u$ with respect to $x$ is actually $\frac{\partial_x u}{dx}$. Now, let's multiply both sides by $\frac{dy}{dx}$:

$$\frac{\partial_x u}{\partial_y u} \frac{dy}{dx} = -\frac{dy}{dx} $$

We can rearrange the left-hand side so that we get partial derivatives:

$$\frac{\frac{\partial_x u}{dx}}{\frac{\partial_y u}{dy}} = -\frac{dy}{dx} $$

And that is the procedure that your friend gave you. However, for a general technique for implicit differentiation, the method that @MrYouMath above gives is generally more straightforward.

$\endgroup$
  • $\begingroup$ By the way, the reason why @MrYouMath's method is preferable is that it works for any number of variables, while this method I believe only works for systems with two variables. $\endgroup$ – johnnyb Feb 27 '18 at 19:35
  • $\begingroup$ I don’t really understand why $du=\delta_x u+ \delta_y u$. Could you please explain this for me. If there is three variables, is the relation $du= \delta_x u + \delta_y u + \delta_z u$ true? $\endgroup$ – Superfrankie Mar 2 '18 at 10:49
  • $\begingroup$ Yes, a total derivative is merely the sum of its partials. Unfortunately, modern notation for differentials makes this horrendously unclear. I don't know if you have done partial differentials yet or not, but the notation is awful. $\endgroup$ – johnnyb Mar 2 '18 at 13:10
  • $\begingroup$ You may learn something like the partial derivative of $u$ with respect to $x$ is $\frac{\partial u}{\partial x}$ This is a horrendously awful notation. A clearer notation is that the partial derivative of $u$ with respect to $x$ is $\frac{\partial_x u}{dx}$. I'm working on a paper to this end. However, if you go with that, every partial derivative formula will start making a lot more sense. $\endgroup$ – johnnyb Mar 2 '18 at 13:12
  • $\begingroup$ Thank you! Thank you for helping me! $\endgroup$ – Superfrankie Mar 2 '18 at 13:34
1
$\begingroup$

You do not have to rearrange the equation just apply the total differential to the equation

$$x^2+y^2=1 \implies d(x^2)+d(y^2)=d(1) \implies 2xdx+2ydy = 0.$$

Now, solve for $dy/dx$.

$\endgroup$
1
$\begingroup$

It is quite reasonable. Note: $$x^2+y^2=1 \Rightarrow w(x,y)=x^2+y^2-1=0.$$ The total differential of the function $w(x,y)$ is: $$w_xdx+w_ydy=0 \Rightarrow \frac{dy}{dx}=-\frac{w_x}{w_y}=-\frac{\frac{dw}{dx}}{\frac{dw}{dy}}.$$ So your classmate is only missing minus.

$\endgroup$
1
$\begingroup$

Similar to the other answers, I would differentiate with respect to $x$ and treat $y=y(x)$ by computing \begin{align} \frac{d}{dx}\left(x^2+y(x)^2\right) &= \frac{d}{dx}(1)\\ \implies 2x+2y\frac{dy}{dx} &= 0\\ \implies \frac{dy}{dx} &= -\frac{x}{y} \qquad \blacksquare \end{align}

$\endgroup$
  • 1
    $\begingroup$ It's not really similar to the others. It's much simper (and better). $\endgroup$ – zhw. Feb 27 '18 at 23:20
0
$\begingroup$

Yes it seems completely unresonable and meaningless.

Indeed the following application of chain rule

$$\frac{dy}{dx}=\frac{dy}{dw}\frac{dw}{dx}$$

has a meaning for $y=f(w(x))$, that is not the case for the definition given for $w$.

Then let use the standard method

$$2xdx+2ydy=0 \implies \frac{dy}{dx}=-\frac{x}{y}$$

$\endgroup$
  • $\begingroup$ I edited my answer, I am sorry that I did not express my self well. $\endgroup$ – Superfrankie Feb 27 '18 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.