Sign up ×
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'm going through the MIT lecture on implicit differentiation, and the first two steps are shown below, taking the derivative of both sides:

$$x^2 + y^2 = 1$$ $$\frac{d}{dx} x^2 + \frac{d}{dx} y^2 = \frac{d}{dx} 1$$ $$2x + \frac{d}{dx}y^2 = 0$$

That makes some sense, but what about this example:

$$x = 5$$ $$\frac{d}{dx} x = \frac{d}{dx} 5$$ $$1 = 0$$

Why is the first example correct, while the second is obviously wrong?

share|cite|improve this question
What is the function $y$ defined implicitly by $x = 5$? – Willie Wong May 31 '13 at 16:28
Christian what do you mean? – Jon Jun 1 '13 at 3:06
Wow! I also saw the video but I never thought so much about it ... How/Where did you get the second example? – Kartik Oct 16 at 14:26

5 Answers 5

up vote 6 down vote accepted

The first of your identities makes some implicit assumptions: it should be read as $$ x^2+f(x)^2=1 $$ where $f$ is some (as yet undetermined) function. If we assume $f$ to be differentiable, then we can differentiate both sides: $$ 2x+2f(x)f'(x)=0 $$ because the assumption is that the function $g$ defined by $g(x)=x^2+f(x)^2$ is constant.

From this we can derive $$ f'(x)=-\frac{x}{f(x)} $$ at least in the points where $f(x)\ne0$, which excludes $x=1$ and $x=-1$ from the domain where $f$ is differentiable.

Thus what you get is that assuming $f$ exists and is differentiable, then, for $x\ne1$ and $x\ne 1$, $f'$ satisfies the above relation.

Why is the relation written in that way? The answer is that often we're given a locus defined by some equation in two variables: it's the set of points $(x,y)$ such that $h(x,y)=0$ and we try finding an explicit form for the locus, that is a relation $y=f(x)$ or $x=g(y)$ , so that $$ h(x,f(x))=0\qquad\text{ or }\qquad h(g(y),y)=0 $$ holds for $x$ in a suitable neighborhood of $x_0$ or $y$ in a suitable neighborhood of $y_0$ where $(x_0,y_0)$ belongs to the locus.

Take for example the folium Cartesii, $x^3+y^3-3xy=0$. If we differentiate with respect to $x$, we get $$ 3x^2+3y^2y'-3y-3xy'=0 $$ which gives $$ y'=\frac{y-x^2}{y^2-x} $$ We're able to find where the derivative is zero by setting $y=x^2$ and plugging in the original equation $$ x^3+x^6-3x^3=0 $$ that is $x=0$ (which can't be used) or $x^3=2$, without even knowing the “explicit form“ $y=f(x)$.

share|cite|improve this answer

$x=5$ implies that $x$ doesn't change so it's meaningless to try to take the derivative of it with respect to $x$

share|cite|improve this answer

You wrote "$x = 5$"; what does that tell us about $x$? Just that, $x$ equals 5. So in differentiating both sides you must keep that in mind. In other words, $x$ is constant and 5 is constant.

Also, then you can't do

$${d \over dx} x = {d \over dx} 5, \tag{1}$$

since that's equivalent to

$${d \over d5} x = {d \over d5} 5, \tag{2}$$

which already has been pointed out is meaningless.

Though you can do

$${d \over dy} x = {d \over dy} 5 \Leftrightarrow 0 =0;\tag{3}$$

here $y$ is an independent variable over the real numbers.

share|cite|improve this answer

$$2x + \frac{d}{dx}y^2 = constant $$

$$ x + y \frac{dy}{dx} = 0$$

That makes sense to seek variation between $x$ and $y$ in terms of differentials for a certain curve, the circle.

Now attacking a constant.

$$x = 5$$

There is no variation it is so clearly known and stated and understood, knowing fully well that fact, we want still to press on to seek a variation between $x$ and $y$ !, just to see what will happen... That makes no sense with such a process... expectedly leading us nowhere,

$$ 1= 0. $$

share|cite|improve this answer

That's because in the first case we can consider an infinitesimal $\mathrm{d}x$ since

$-1 \lt x \lt 1$. However for a constant $x$ in the second case the $\mathrm{d}x$ is meaningless (can consider it always $\mathrm{d}x = 0$). So $\frac{0}{0} = \frac{\mathrm{d}x}{\mathrm{d}x} = \frac{1}{\mathrm{d}x} = \frac{1}{0}$ !!?

share|cite|improve this answer

Your Answer


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.