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I was just wondering where the y'/(dy/dx) in implicit differentiation comes from. $$ x^2 + y^2 = 25 $$ $$ (d/dx) x^2 + (d/dy) y^2 **(dy/dx)** = 25 (d/dx) $$ $$ 2x + 2y (dy/dx) = 0 $$ $$ (dy/dx) = -x/y $$ Where does the bold part come from? Wikipedia says it's a byproduct of the chain rule, but it's just not clicking for me.

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    $\begingroup$ When you do implicit differentiation what you're doing is assuming $y(x)$ (that $y$ is a function of $x$). Then you're viewing the equation $x^2+y^2=25$ as an equality between functions of $x$ -- it's just that the right-hand side is the constant function $25$. So you differentiate the left and right-hand sides. The derivative of $y^2$ with respect to $x$ is $2yy'$ by the chain rule. $\endgroup$ Commented Oct 24, 2010 at 20:16
  • $\begingroup$ Where does the bold part go to? $\endgroup$
    – miracle173
    Commented Apr 18, 2012 at 18:12
  • $\begingroup$ It's the part with the *****(dy/dx)***** around it. Someone has altered the formatting $\endgroup$
    – Ryan
    Commented Apr 19, 2012 at 1:04

2 Answers 2

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When you implicitly differentiate $x^2+y^2=25$, you are differentiating with respect to a particular variable—in this case, $x$, so: $$\begin{align} \frac{d}{dx}(x^2+y^2)&=\frac{d}{dx}25 \\ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)&=0 \\ 2x+2y\frac{dy}{dx}&=0 \\ 2y\frac{dy}{dx}&=-2x \\ \frac{dy}{dx}&=-\frac{x}{y} \end{align}$$

From the 3rd line to the 4th line, $\frac{d}{dx}(y^2)$ is the derivative with respect to $x$ of $y^2$, in which (as in Ryan Budney's comment) we assume that $y$ is some function of $x$, so we apply the chain rule, differentiating $y^2$ with respect to $y$ and multiplying by the derivative of $y$ with respect to $x$ to get $2y\frac{dy}{dx}$.


edit: Based on the comments below, I think it might be useful if I introduced a slightly different notation: Let $D_x$ be the differential operator with respect to $x$, which you have previously written as $\frac{d}{dx}$ (and, similarly, $D_y$ is the differential operator with respect to $y$). When we apply the differential operator to something, we read and write it like a function: $D_x(x^2)=2x$ is "the derivative with respect to $x$ of $x^2$ is $2x$."

Now, rewriting the work above in this notation:

$$\begin{align} D_x(x^2+y^2)&=D_x(25) \\ D_x(x^2)+D_x(y^2)&=0 \\ 2x+D_y(y^2)D_x(y)&=0 \\ 2x+2yD_x(y)&=0 \\ 2yD_x(y)&=-2x \\ D_x(y)=\frac{dy}{dx}&=-\frac{x}{y} \end{align}$$

And, to your question of finding $\frac{dx}{dy}$: $$\begin{align} D_y(x^2+y^2)&=D_y(25) \\ D_y(x^2)+D_y(y^2)&=0 \\ D_x(x^2)D_y(x)+2y&=0 \\ 2xD_y(x)+2y&=0 \\ 2xD_y(x)&=-2y \\ D_y(x)=\frac{dx}{dy}&=-\frac{y}{x} \end{align}$$

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  • $\begingroup$ So, would (d/dy) x^2 (dx/dy) + (d/dx) y^2 = 25 (d/dy) --> 2x (dx/dy) + 2y = 0 --> (dy/dx) = -y/x be an equally valid answer? $\endgroup$
    – Ryan
    Commented Oct 25, 2010 at 1:24
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    $\begingroup$ @Ryan: Sort of. It's correct in this setting (I think $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$ is true when dealing with a curve in the x-y plane) but it's not really "proper" in that you should not treat $\frac{dy}{dx}$ and $\frac{dx}{dy}$ as fractions that are reciprocals of one another. That is, if you were a student of mine in a first-year calculus course, I'd give you most or all of the credit, but you'd also get a note that you shouldn't do it that way and should differentiate with respect to the correct variable. $\endgroup$
    – Isaac
    Commented Oct 25, 2010 at 2:56
  • $\begingroup$ Sorry, what I had meant was if they had asked you to solve for dx/dy, would what I put there have been correct? $\endgroup$
    – Ryan
    Commented Oct 25, 2010 at 4:17
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    $\begingroup$ @Ryan: ahh... In the first step, I think you have (d/dy) and (d/dx) swapped on the left side and the 25 should be after d/dy on the right side, and in the last step, I think you meant dx/dy. That is, I think it should be: ((d/dx) (x^2)) (dx/dy) + (d/dy)(y^2) = (d/dy)(25) --> 2x (dx/dy) + 2y = 0 --> (dx/dy) = -y/x $\endgroup$
    – Isaac
    Commented Oct 25, 2010 at 4:25
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    $\begingroup$ @Ryan: Written in the notation I introduced, the chain rule is more or less Dx(whatever)=Dy(whatever)Dx(y). In evaluating Dx(y^2), we're working as if y is a function of x, so by the chain rule, we can differentiate y^2 with respect to y and multiply by the derivative of y with respect to x. Put another way, the chain rule is often expressed as: if h(x)=f(g(x)), then h'(x)=f'(g(x))g'(x); in the case of Dx(y^2), f(y)=y^2 and g(x)=y(x), so h(x)=(y(x))^2 and h'(x)=f'(g(x))g'(x)=2(y(x))y'(x). $\endgroup$
    – Isaac
    Commented Oct 25, 2010 at 4:51
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Isaac and Ryan have already answered your question in words. Now, in symbols, the chain rule gives: $$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\frac{dy}{dx} = 2y\frac{dy}{dx}$$

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  • $\begingroup$ +1: This is why I had the persistent feeling that my answer was missing something. $\endgroup$
    – Isaac
    Commented Oct 24, 2010 at 22:14
  • $\begingroup$ i never understood this form...and it causes all sorts of problem. The following make sense..d/dx f(g(x))= f'(g(x)) g'(x). But in this form it dosen't: ..= d/dy f(g(x) (d/dx g(x)) the part d/dy dosen't make sense to me i thought it was just a notation showing that derivative/Instantaneous slope of y,c,etc is this as x changes. $\endgroup$ Commented Dec 4, 2013 at 5:36

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