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I have seen how people implicitly differentiate the equation $x^2 + y^2 = c$.

$$d/dx(x^2) + d/dx(y^2) = d/dx(c)$$

treating "$y$" as "$f(x)$" and using the chainrule we get

$$2x + 2y(y') = 0$$

and solving for $y'$

$$y'= -2x/2y$$

The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both "$x$" and "$y$" on the same side of the equation is what´s bothering me and even if I see "$y$" as a function of "$x$" I just end up imagining

$$x^2 + (-x^2 + c) = c$$

which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

I really appreciate any thoughts or ideas. Thank you!

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  • 2
    $\begingroup$ Possible duplicate of Implicit differentiation $\endgroup$ – Git Gud Aug 17 '16 at 21:14
  • $\begingroup$ Please format your question properly. Implicit differentiation is best understood by knowing what physics the Chain Rule is supposed to model. The following question should help. If you have a ball of ice, which melts so that the radius shrinks at the rate $r'(t) = m(t)$, at what rate does the volume shrink? $\endgroup$ – avs Aug 17 '16 at 21:15
  • $\begingroup$ @GitGud: The answer provided there might be too technical for the current situation. $\endgroup$ – user66081 Aug 17 '16 at 21:20
  • $\begingroup$ @user66081 "The" isn't right. There are several answers. Are you of the opinion that all of them are too technical? I did consider not closing as a duplicate, but since my vote is just one fifth of the closing contribution, I decided to vote to close. $\endgroup$ – Git Gud Aug 17 '16 at 21:22
  • $\begingroup$ @Leo (cc. @GitGud): maybe you could comment if/why you find the answers to the previous question unhelpful. That would be of general interest. $\endgroup$ – user66081 Aug 17 '16 at 21:26
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Maybe this helps. Take the function $f(x) = x^2$ and some other functions $g(x)$ and $h(x)$.

Let's differentiate the expression $$ f(x) + (g(x))^2 = h(x) $$ with respect to $x$ (not: "differentiate $x$").

We use the fact that you can differentiate each summand individually, and the chain rule for $(g(x))^2$, to get $$ f'(x) + 2 g(x) g'(x) = h'(x) . $$

In your case, $f(x) = x^2$, $g(x) = y(x)$, and $h(x) = c$, i.e.: $$ 2 x + 2 y(x) y'(x) = 0 . $$

Sometimes, the argument $(x)$ is omitted -- that's what you have there; but that's just notation (no deeper meaning, afaik).

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  • $\begingroup$ Now I get. Thank you so much! That was actually a really good explanation, at least for me. $\endgroup$ – Leo Aug 17 '16 at 21:26
  • $\begingroup$ @Leo: Ok. Feel free to accept one of the answers. I think even better would be if you could write your own answer and explain the difficulties. $\endgroup$ – user66081 Aug 17 '16 at 21:29
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We are just using the chain rule

If I said:

$f(x) = (2x +1) \implies \frac {df}{dx} = 2\\ g(x) = x^2\implies \frac {dg}{dx} = 2x\\ \frac {d}{dx}(g\circ f)(x) = \frac {dg}{df}\frac {df}{dx} = 2(2x+1)(2)$

Do have a problem with that?

So, $y$ is a function of $x.$ $y^2$ is a function of $y$

$\frac {d}{dx} y^2 = \frac {d}{dy}(y^2)\frac {dy}{dx}=2y\frac {dy}{dx}$

And now for the last little bit:

$\frac {d}{dx} (y^2 = c - x^2) \implies 2y\frac {dy}{dx} = -2x$

Is that so different from

$\frac {d}{dx} (x^2 + y^2 = c)\\ \frac {d}{dx} (x^2) + \frac {d}{dx} (y^2) = \frac {d}{dx} c \\ 2x + 2y\frac {dy}{dx} = 0\\ \frac{dy}{dx} = -\frac {x}{y}$

Does this help?

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  • $\begingroup$ Please, please, please do not write $dg/df$. $\endgroup$ – Ted Shifrin Aug 17 '16 at 21:36
  • $\begingroup$ @TedShifrin why not? $\endgroup$ – Doug M Aug 17 '16 at 21:36
  • $\begingroup$ Because $g$ is not a function of the variable $f$. $\endgroup$ – Ted Shifrin Aug 18 '16 at 1:15
  • $\begingroup$ In the example above? Sure it is. It is a composition. g(f(x)) $\endgroup$ – Doug M Aug 18 '16 at 1:20
  • $\begingroup$ No, that's not what function means. You mean $\dfrac{dg}{dx}(f(x))$, since you gave $g$ as a function of $x$ to start with. It would be far clearer to give $g(u)=u^2$ and write $\dfrac{dg}{du}\big|_{u=f(x)}$. $\endgroup$ – Ted Shifrin Aug 18 '16 at 1:24
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When you differentiate, you differentiate with respect to a variable. As you remember, a derivative $dx/dt$ essentially means how much does x change when t is changed by an extremely small amount.

Recall the chain rule, which is the rule for differentiating composite functions: $$ \frac{d}{dx}f(g(x)) = f'(g(x))g'(x) $$

In this case, you're differentiating with respect to $x$.

So for the $x^2$ term you can view this as the composition of these two functions $f(x) = x^2$ and $g(x) = x$. Applying the chain rule, you get $$ \frac{d}{dx}x^2 = \frac{d}{dx}x^2(\frac{d}{dx}x) = 2x $$

Now we get to the $y^2$ term. For this, your functions are $f(x) = x^2$ and $g(x) = y$. Again, applying the chain rule gives $$ \frac{d}{dx}y^2 = \frac{d}{dx}x^2(\frac{d}{dx}y) = 2x(y') $$

We have no information on the relationship between y and x, so all we can write is $\frac{dy}{dx}$ or $y'$ for short.

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