If $z= \arctan \frac{y}{x}$ show that the following is true $$x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0$$

So I don't truly understand how implicit partial differentiation works, but I understand normal implicit differentiation. I would be great if someone would be able to give some help with this.

The question also gives a side note of $$\frac{\partial}{\partial x}\left( \arctan z \right) = \frac{1}{1+z^2}\times \frac{\partial z}{\partial x}$$

Im thinking you start by putting the euqation as $\tan z =\frac{y}{x}$?

  • 3
    $\begingroup$ Implicit differentiation is unnecessary here, you only need the two partial derivatives of z, which is a function of x and y. So just differentiate z wrt x (i.e. treat y as constant), and differentiate z wrt y, and the required identity should follow. If you're struggling with partial differentiation, say wrt x, then it may help to view y as a concrete constant like 2. $\endgroup$ – AlephNull Jan 1 at 16:27
  • $\begingroup$ So it can be done with and without implict? $\endgroup$ – H.Linkhorn Jan 1 at 16:28
  • $\begingroup$ Yes, provided that you know the derivative of arctan, which is given implicitly in the side note you mentioned anyway. (When I say implicitly here I don't mean it in the mathematical sense! Edit: Though I suppose it is also true in the mathematical sense.) $\endgroup$ – AlephNull Jan 1 at 16:30

Just for the fun of it, and for the sake of "completeness", let's first find the derivative of $\arctan w$ ourselves:


$z = \arctan w; \tag 1$


$w(z) = \tan z = \dfrac{\sin z}{\cos z}; \tag 2$

we have

$w'(z) = \dfrac{(\cos z)(\cos z) - (-\sin z)(\sin z)}{\cos^2 z} = \dfrac{\cos^2 z + \sin^2 z}{\cos^2 z} = \dfrac{1}{\cos^2 z} = \sec^2 z; \tag 3$ we use the identity

$1 + \tan^2 z = 1 + \dfrac{\sin^2 z}{\cos^2 z} = \dfrac{\cos^2 z + \sin^2 z}{\cos^2 z} = \dfrac{1}{\cos^2 z} = \sec^2 z \tag 4$

to write

$\dfrac{dw}{dz} = w'(z) = 1 + \tan^2 z = 1 + w^2, \tag 5$


$z'(w) = \dfrac{dz}{dw} = \dfrac{d(\arctan w)}{dw} = \dfrac{1}{1 + w^2}. \tag 6$

Now having $z'(w)$ at hand, we set

$w = \dfrac{y}{x} = yx^{-1}, \tag 7$

and use the chain rule to find $\partial z / \partial x$, $\partial z / \partial y$:

$\dfrac{\partial z}{\partial x} = \dfrac{dz}{dw} \dfrac{\partial w}{\partial x}, \tag 8$

$\dfrac{\partial w}{\partial x} = \dfrac{\partial (yx^{-1})}{\partial x} = -yx^{-2}; \tag 9$

we fold (6) and (9) into (8):

$\dfrac{\partial z}{\partial x} = -yx^{-2}\dfrac{1}{1 + w^2} , \tag{10}$


$\dfrac{\partial w}{\partial y} = \dfrac{\partial (yx^{-1})}{\partial y} = x^{-1}, \tag{10}$


$\dfrac{\partial z}{\partial y} = \dfrac{dz}{dw} \dfrac{\partial w}{\partial y} = x^{-1}\dfrac{1}{1 + w^2}; \tag{11}$


$x \dfrac{\partial z}{\partial x} + y \dfrac{\partial z}{\partial y} = -yx^{-1}\dfrac{1}{1 + w^2} + yx^{-1}\dfrac{1}{1 + w^2} = 0, \tag{12}$

as was to be proved.

So much for the "standard derivation" based upon straightforward partial differentiation. However,

There is in fact a much swifter, easier way to do this:

We have the radial vector

$\mathbf r = \begin{pmatrix} x \\ y \end{pmatrix}, \tag{13}$

and that the central angle which $\mathbf r$ makes with the $x$-axis is

$\theta = \arctan \dfrac{y}{x} = z; \tag{14}$

it follows that

$\nabla \theta = \begin{pmatrix} \dfrac{\partial \theta}{\partial x} \\ \dfrac{\partial \theta}{\partial y} \end{pmatrix} = \begin{pmatrix} \dfrac{\partial z}{\partial x} \\ \dfrac{\partial z}{\partial y} \end{pmatrix}; \tag{15}$


$\mathbf r \cdot \nabla \theta = \nabla_{\mathbf r} \theta = 0, \tag{16}$

since $\theta$ does not change in the $\mathbf r$ direction; indeed, we see that, for $0 < \alpha \in \Bbb R$,

$\arctan \dfrac{\alpha y}{\alpha x} = \arctan \dfrac{y}{x}, \tag{17}$

which shows that $\theta$ is invariant along the rays $\alpha \mathbf r$. Therefore,

$x \dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} = \mathbf r \cdot \nabla \theta = \nabla_{\mathbf r} \theta = 0, \tag{18}$

as required. $OE\Delta$.


Let $u=y/x$ so $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=x\cdot\frac{-y}{x^2}+y\cdot\frac{1}{x}=0$. Then $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{dz}{du}\cdot 0=0$.


$$\frac{\partial z}{\partial x}=\frac{y}{1+\frac{y^2}{x^2}}\cdot (\frac{-1}{x^2})=\frac{-y}{x^2+y^2}$$


$$\frac{\partial z}{\partial y}=\frac{1}{1+\frac{y^2}{x^2}}\cdot \frac1x=\frac{x}{x^2+y^2}.$$

Now substitute in $$x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}.$$


Here is a way to do this without implicit differentiation. $$x\frac{\partial z}{\partial x}=\frac{x}{1+(yx^{-1})^2}\frac{-1}{x^2}=-\frac{1}{x(1+(yx^{-1})^2)}$$ $$y\frac{\partial z}{\partial y}=\frac{x^{-1}}{1+(yx^{-1})^2}=\frac{1}{x(1+(yx^{-1})^2)}$$ Therefore, $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$$


first apply the chain rule to evaluate the partial derivative. $$\frac{\partial z}{\partial x}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(-\frac{y}{x^2}\right)$$ $$\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(\frac{1}{x}\right)$$ so,now $$x\frac{\partial z}{\partial x}=\frac{x}{1+\left(\frac{x}{y}\right)^2}\left(-\frac{y}{x^2}\right)=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(-\frac{y}{x}\right)......(1)$$ $$y\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left(\frac{y}{x}\right)..................(2)$$ adding (1)+(2), $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{x}{y}\right)^2}\left[\left(-\frac{y}{x}\right)+\left(\frac{y}{x}\right)\right]=0$$


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