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????

  • $\begingroup$ 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

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.

  • $\begingroup$ 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? $\endgroup$ – Siwei 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! $\endgroup$ – Siwei Mar 2 '18 at 13:34

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 $dy/dx$.


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.


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}

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

Yes it seems completely unresonable and meaningless.

Indeed the following application of chain rule


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

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

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

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