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On trying to find $\displaystyle{\frac{dy}{dx}}$ when the given equation is $\displaystyle{xy = \cot{xy}}$, the following step gives an equation which gets rid of the derivative as: $\displaystyle{\frac{d}{dx}(xy) = - \csc^{2}{(xy)} \frac{d}{dx}(xy) }$

what's more, it gives an equation with no $(x,y)$ in real plane satisfying the equation, as: $\displaystyle{\csc^{2}{(xy)} = -1}$

To be sure that the function isn't actually differentiable, I graphed it on desmos, it looks although discrete at some points away from origin but I take it as inability of the graphing software to plot precisely, also supported by the seeming smoothness of the curve near origin.

So should I consider it non differentiable and leave the problem there or is there just any mistake i did while differentiating, and the curve is actually differentiable?

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    $\begingroup$ The posted answers give the solution. Your mistake was in overlooking the solution $\frac{d}{dx}(xy)=0$. $\endgroup$ – Teepeemm Feb 15 at 1:33
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The equation $$t=\cot t$$ has infinitely many solutions, let $t_k$.

So

$$xy=\cot xy$$ describes infinitely many equilateral hyperbolas

$$xy=t_k.$$

Every branch is differentiable ($\dfrac{dy}{dx}=-\dfrac{t_k}{x^2}$), but collectively, $y$ is not a univocal function of $x$.


If you want to use implicit differentiation,

$$xy=\cot xy$$ yields

$$y+xy'=(-\csc^2xy)\,(y+xy')$$ which is

$$y'=-\frac yx=-\frac{xy}{x^2}.$$

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\begin{align} xy&=\cot xy\\ \frac{d}{dx}\left(xy\right)&=\frac{d}{dx}\left(\cot xy\right)\\ y+x\frac{dy}{dx}&=-\csc^2xy\cdot\left(y+x\frac{dy}{dx}\right)\\ \left(1+\csc^2xy\right)\left(y+x\frac{dy}{dx}\right)&=0\\ \frac{dy}{dx}&=-\frac{y}{x} \end{align}

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First, are there any solutions to $x y = \cot(x y)$?

Plot of t and cot(t)

Each intersection corresponds to a value of $x y$ for which $x y = \cot(x y)$. Numerically, the intersections in that plot occur for $x y \in \{-9.52933{\dots}, -6.4373{\dots}, -3.42562{\dots}, -0.860334{\dots}, 0.860334{\dots}, 3.42562{\dots}, 6.4373{\dots}, 9.52933{\dots} \}$. The intersections continue to the left and right asymptotically approaching integer multiples of $\pi$.

Suppose $k$ is a value of $x y$ such that $x y = \cot(x y)$. What is the locus of points satisfying $x y = k$? It is a hyperbola, $y = k/x$. The particular solutions numerically found above correspond to these hyperbolae.

Hyperbolae

Each hyperbola is two arcs (either one in QI and one in QIII or one in each of QII and QIV) and in the plot, both arcs have the same color. Further positive values of $xy$ giving intersections in the first graph give more hyperbolae (in QI and QIII). Further negative values give more hyperbolae (in QII and QIV).

Notice that on each hyperbola, the quantity $x y$ is constant. That is, $\frac{\mathrm{d}(xy)}{\mathrm{d}x} = 0$. So your equation $$ \frac{\mathrm{d}}{\mathrm{d}x}(xy) = - \csc^{2}{(xy)} \frac{\mathrm{d}}{\mathrm{d}x}(xy) $$ has another solution you are not thinking of. When you divide both sides by $\frac{\mathrm{d}}{\mathrm{d}x}(xy)$, you lose this solution. The manipulation that preserves all solutions is \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(xy) &= - \csc^{2}{(xy)} \frac{\mathrm{d}}{\mathrm{d}x}(xy) \\ \frac{\mathrm{d}}{\mathrm{d}x}(xy) + \csc^{2}{(xy)} \frac{\mathrm{d}}{\mathrm{d}x}(xy) &= 0 \\ \left( \frac{\mathrm{d}}{\mathrm{d}x}(xy) \right) \left(1 + \csc^{2}{(xy)} \right) &= 0 \\ \end{align*} Since the product of quantities is zero when one (or more) of them are zero, we have a solution when either $\frac{\mathrm{d}}{\mathrm{d}x}(xy) = 0$ or $1 + \csc^{2}{(xy)} = 0$ (or both). We have already shown the hyperbolae of $x$-$y$ pairs that satisfy the first of these. The second has no solution: since $\csc^{2}{(xy)} \geq 0$ as long as the cosecant is defined, $1 + \csc^{2}{(xy)} \geq 1 > 0$.

This says the graph of the original implicit equation is the infinite set of hyperbolae (of which we graphed eight). As others have shown, if you complete your implicit differentiation (remembering to use the chain rule for $y$, since we treat it as if it were $y(x)$, a function of $x$), you will see that the slope of a tangent line to the point $(x,y)$ on one of those hyperbolae has slope $-y/x$.

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