What happens when we take the derivative of an implicitly defined function?

Question

Consider the function y(x) implicitly decided by this equation: $\sin(xy)=\cos(x+y)$.

The graph on plane looks like this:

According to my textbook, we can take the derivative of both sides with respect to x to obtain the derivative y'(x)=[some function of x and y].

In this case y'(x) = -(\sin(xy)+y\cos(xy))/(\sin(x+y)+x\cos(xy)).

In this case $y'(x) = -\dfrac{\sin(x+y)+y\cos xy }{\sin(x+y)+x\cos xy }$.

Now evaluate $y'(x)$ at the point $A$. There seems to be two tangent lines at point $A$. Then which slope does $y'(x)$ at $A$ describe?

Answer 1

You've solved for $\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y)$, but the function you $f$ you have is not defined everywhere. Indeed, at those points of intersection $(x_0,y_0)$ where there is more than one tangent line, if you try to plug $(x_0,y_0)$ into $f$ you get $0/0$ which is undefined. Instead, you have to take the limit of $f(x,y)$ as $(x,y)\to(x_0,y_0)$ along the appropriate portion of $f$'s graph.

A simpler example is $y^2=x^2$ (which is equivalent to $y=\pm x$, which is just a big X). Differentiating and solving yields $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x}{y}$. Along the path $y=x$ the ratio $\frac{x}{y}$ is identically $1$ so the limit is $1$, whereas on the path $y=-x$ the ratio is identically $-1$ so the limit is $-1$. As you can see by looking at the X, the two possible slopes are indeed $+1$ and $-1$.

Answer 2

Here is a graph of the implicit function $y$ along with the graphs of the set where the numerator of $y'$ vanishes and the set where the denominator of $y'$ vanishes. Note they all meet at $A$, so your expression for $y'$ becomes $0/0$ at $A$. That means you can't use your expression for $y'$ at the point $A$.

Answer 3

It may be of interest to note that the self-intersection points all lie on the lines $ \ y \ = \ n \pi - x \ \ $ or $ \ x + y \ = \ n \pi \ \ . \ $ [The red diagonal line is $ \ x + y \ = \ 0 \ \ , \ $ the orange lines, $ \ x + y \ = \ \pm \pi \ \ , \ $ the green lines, $ \ x + y \ = \ \pm 2\pi \ \ . \ ] \ $ So at these points, we have $ \ \sin ( \ x·[n \pi - x] \ ) \ = \ \pm 1 \ \ , \ $ with the sign alternating on successive lines (at your point $ \ A \ \ , \ \sin ( \ x·[- \pi - x] \ ) \ = \ -1 \ \ ) \ . \ $

The derivative at these points is then $$ y' \ \ = \ \ \frac{0 \ + \ (n \pi \ - \ x)·\cos( \ x·[ n \pi \ - \ x ] \ )}{0 \ + \ x·\cos( \ x·[ n \pi \ - \ x ] \ )} \ \ = \ \ \frac{ (n \pi \ - \ x)·\cos( \ x·[ n \pi \ - \ x ] \ )}{x·\cos( \ x·[ n \pi \ - \ x ] \ )} \ \ . $$ Since we have $ \ \sin ( \ x·[- \pi - x] \ ) \ = \ \pm 1 \ $ there, the cosine factors are zero, whence the indeterminate ratio for $ \ y' \ \ (x \ \neq \ 0 \ $ at any of those points).

Why are there two curves crossing through point $ \ A \ $ (or any of these other points)? We can apply the "angle-addition" identities to the two sides of the curve equation to gain (perhaps) some insight. On the line $ \ y \ = \ -\pi - x \ \ , \ $ the left side is $ \ \sin(xy) \ = \ \sin ( \ x·[- \pi - x] \ ) \ \ , \ $ which becomes

$$ \sin ( \ - \pi·x \ - \ x^2 \ ) \ \ = \ \ -\sin ( \ \pi·x \ + \ x^2 \ ) \ \ = \ \ -[ \ \sin (\pi·x)·\cos(x^2) \ + \ \cos (\pi·x)·\sin(x^2) \ ] \ \ , $$

equal to $ \ \cos (- \pi) \ = \ -1 \ \ $ on the right side. Introducing a small change in the value of $ \ \cos(x + y) \ = \ -1 + \varepsilon \ $ leads to two admissible values for $ \ x \ \ $ (and so two values each for $ \ y \ ) \ \ , \ $ indicating that there can be two sections of the curve along which points in the neighborhood give values close to the minimum possible for $ \ \sin ( \ - \pi·x \ - \ x^2 \ ) \ \ . \ $ We see this when we examine the values of this expression along the portions of the curve passing through $ \ A \ ( \ \approx \ -3.58 \ , \ \approx \ 0.439 \ ) \ \ , \ $ presented in the graph below. (I don't know what a "good" parameterization for this curve would be.)

The easiest conclusion to reach for this curve is that the self-intersections occur then at minima and maxima of $ \ \sin(xy) \ $ and $ \ \cos(x + y) \ \ , \ $ through which two "trajectories" pass at each location in different directions. When this happens, there will be more than one value for $ \ \frac{dy}{dx} \ \ , \ $ which the expression from implicit differentiation will not resolve. We generally need a curve parameterization or some other sort of detailed analysis of the curve relation to determine what the (limiting) values of the multiple slopes are.