Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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. –  Ryan Budney Oct 24 '10 at 20:16
    
Where does the bold part go to? –  miracle173 Apr 18 '12 at 18:12
    
It's the part with the *****(dy/dx)***** around it. Someone has altered the formatting –  Ryan Apr 19 '12 at 1:04

2 Answers 2

up vote 5 down vote accepted

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

share|improve this answer
    
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? –  Ryan Oct 25 '10 at 1:24
1  
@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. –  Isaac Oct 25 '10 at 2:56
    
Sorry, what I had meant was if they had asked you to solve for dx/dy, would what I put there have been correct? –  Ryan Oct 25 '10 at 4:17
1  
@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 –  Isaac Oct 25 '10 at 4:25
1  
@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). –  Isaac Oct 25 '10 at 4:51

Isaac and Ryan have already answered your question in words. Now, in symbols, the chain rule gives: $$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\frac{dy}{dx} = 2y\frac{dy}{dx}$$

share|improve this answer
    
+1: This is why I had the persistent feeling that my answer was missing something. –  Isaac Oct 24 '10 at 22:14
    
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. –  Muhammad Umer Dec 4 '13 at 5:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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