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I want to differentiate

$x^2 + y^2=1$ with respect to $x$.

The answer is $2x +2yy' = 0$.

Can some explain what is implicit differentiation and from where did $y'$ appear ? I can understand that $2x +2yy' = 0$ is a partial derivative but then it becomes multi calc not single.

This is in a chapter about chain rule so I assume there is some use of the chain rule here but I can't see any composite functions here. We can express y in terms of x but y is not composite.

P.S: I am NOT looking for HOW to solve the problem, I am looking for the WHY as stated above.

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1  
do you mean you differentiate with respect to $x$? –  user88595 Nov 16 '13 at 23:49
    
yeah sorry for the typo –  Nameless Nov 16 '13 at 23:50
    
Treat $y$ as a function of $x$. Then when you differentiate $x \mapsto y(x)^2$ you will get $x \mapsto 2 y(x) y'(x)$. –  copper.hat Nov 16 '13 at 23:52
    
Related (duplicate?): math.stackexchange.com/questions/407822/… –  egreg Nov 17 '13 at 0:01
    
@egred I read the one you posted before posting this question, but didn't find the answer, as I was wondering how to differentiate an equation as "gitgud" pointed out. –  Nameless Nov 17 '13 at 0:04

5 Answers 5

up vote 4 down vote accepted

This is a matter of understanding what you're dealing with.

You're asked to differentiate $x^2+y^2=1$. An equation isn't a differentiable function, therefore the equation can't be differentiated.

Now comes the 'translating the problem part'.

The equation $x^2+y^2=1$ 'defines a function', more precisely, there exists a function $g\colon U\to V$ such that $x^2+(g(x))^2=1$, for some sets $U$ and $V$. (A lot can be said about $g, U$ and $V$). Let's assume for the time being that $g$ is differentiable. Now what the problems is actually asking you to do is to differentiate both sides of $x^2+(g(x))^2=1$, yielding $2x+2g(x)g'(x)=0$.

All this is simply the Implicit Function Theorem. The details can be checked on the link.

In two dimensions the theorem goes as follows:

Let $D\subseteq \Bbb R^2$ be an open set and let $f\colon D \to \Bbb R$ be a class $C^1$ function. Given $a\in \Bbb R$, suppose there exists $(x_0, y_0)\in D$ such that $f(x_0, y_0)=a$ and $f_y(x_0, y_0)\neq 0$. Then there are open intervals $U$ and $V$ with the property that there exists a class $C^1$ function $g\colon U\to V$ such that $\forall x\in U\left(f(x,g(x))=c\right)$. Furthermore defining $h\colon U\to \Bbb R, x\mapsto f(x,g(x))$, the chain rule yields $\forall x\in U(h'(x)=f_x(x,g(x))+f_y(x,g(x))g'(x)=0)$.

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The sets U and V must be open... –  Chris K Nov 17 '13 at 0:01
    
@ChrisK Yes, your point? –  Git Gud Nov 17 '13 at 0:01
    
Just a matter of being mathematically precise. I guess it's not necessary to mention it. You can take it or leave it. –  Chris K Nov 17 '13 at 0:03
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@ChrisK I did say a lot could be said about $g, U$ and $V$. –  Git Gud Nov 17 '13 at 0:04
    
You've made a good point. –  Chris K Nov 17 '13 at 0:05

This is called the product rule, not the chain rule.

Differentiating $x^2$ w.r.t $x$ gives $2x$ which can be proven using the definition.

Now, the product rule states that (uv)' = u'v + uv'. So here $y^2 = yy$ ie $(y^2)' = (yy)' = y'y + yy' = 2yy'$ where $y' = \frac{dy}{dx}$

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Actually you can also use the chain rule by stating $f(y) = y^2$ ie $(f(y))' = y'f'(y) = y' \cdot 2y = 2yy'$ –  user88595 Nov 16 '13 at 23:54
    
You're showing that in this case, one can get by without the chain rule, but I think it's pretty important to understand how the chain rule is involved in implicit differentiation. –  Michael Hardy Nov 17 '13 at 0:02

Let $f(x,y) = x^{2} + y^{2} - 1$. Then we claim that a function $g(x)$ satisfying $f(x,g(x)) = 0$ exists. By the implicit function theorem, it suffices to show that $\frac{\partial f}{\partial y}(x, y) = 2y \neq 0$. So the theorem holds and $g(x)$ is differentiable whenever $y\neq 0$. So when solving this problem we're essentially assuming that $y$ can be expressed in terms of $x$. The "obvious" problem is when $x=-1$ or $x=1$, but then $y=0$ and $y'$ is undefined. Now that I've shown $g(x)$ usually exists, we can proceed with chain rule. Since we are on a level set, $D_{1}[f(x,y)] = D_{1}f(x,y) + D_{2}f(x,y)\cdot g'(x)\equiv 0$. So, $g'(x) = \frac{-D_{1}f(x,y)}{D_{2}f(x,y)} = \frac{-2x}{2y} = \frac{-x}{y}$. Now these last steps are exactly what you did by "implicitly differentiating". I must admit that my answer is more of a cautionary tale than anything, but I feel that it is important.

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Cautionary tales mark the difference between physicists and mathematicians. –  Git Gud Nov 17 '13 at 0:15
    
Perhaps this is true. –  Chris K Nov 17 '13 at 0:28
    
I just realised you study physics xD Sorry. –  Git Gud Nov 17 '13 at 0:30

The function $x^2 + y^2 = 1$ defines $y$ implicitly as a function of $x$. In this case, we have $y^2 = 1 - x^2$. Thus, instead or writing $y$ in the equation we can write $f(x)$ where $f(x)^2 = 1 - x^2$. This leaves the problem of differentiating $x^2 + f(x)^2 = 1$. In this form, we can see how to apply the chain rule

$$ \begin{align*} & \frac{d}{dx} \left( x^2 + f(x)^2 \right) && \\ ~ = ~ & \frac{d}{dx} x^2 + \frac{d}{dx} f(x)^2 && [\text{linearity}] \\ ~ = ~ & 2x + \frac{d}{dx} f(x)^2 && [\text{power rule}] \\ ~ = ~ & 2x + 2f(x)\frac{d}{dx}f(x) && [\text{chain rule}] \end{align*}$$

In the fourth line, we can apply the chain rule because $f(x)^2$ is the composition of two functions, namely the function $g(x) = x^2$ and $h(x) = f(x)$. The chain rule says that to differentiate $(g \circ h)(x)$, we first differentiate $g(x)$ with respect to $x$, then differentiate $h(x)$ with respect to $x$, then form the product $(g' \circ h)(x)h'(x)$ as the result. In this case, differentiate $g(x)$ with respect to $x$ gives $2x$. Differentiating $h(x)$ with respect to $x$ gives $h'(x)$. Thus, $(g \circ h)(x) = 2f(x)f'(x)$.

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In this case we consider $ y $ as function of $ x $, so the differential of $ \frac{dy^{2}}{dx}=2y\frac{dy}{dx}$ where $ y^{'}=2y\frac{dy}{dx}$ thus

\begin{align} \dfrac{d\left[ x^{2}+y^{2}\right]}{dx} &= \dfrac{d\left[ x^{2}\right]}{dx}+ \dfrac{d\left[ y^{2}\right]}{dx}\\ &= 2x + \frac{dy^{2}}{dx}=2x+2y\frac{dy}{dx} \\ \left[ x^{2}+y^{2}\right]^{'} &= 2x+ 2yy^{'}. \end{align}

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