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.
Implicit Differentiation
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0It's the part with the *****(dy/dx)***** around it. Someone has altered the formatting – 2012-04-19
2 Answers
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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0sorry...why are we finding derivative of y2 in respect to dy and not dx as in ((d/dy)y^2) <---biggest reason for confusion. – 2013-12-04
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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0i 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. – 2013-12-04