Cfrgtkky

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 $tan z =frac{y}{x}$?

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

Set

$z = arctan w; tag 1$

then

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

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

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

whence

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

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):

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

also,

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

whence,

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

therefore,

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

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

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

thus,

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

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,

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

as required. $OEDelta$.

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

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

and

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

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.
$$xfrac{partial z}{partial x}=frac{x}{1+(yx^{-1})^2}frac{-1}{x^2}=-frac{1}{x(1+(yx^{-1})^2)}$$
$$yfrac{partial z}{partial y}=frac{x^{-1}}{1+(yx^{-1})^2}=frac{1}{x(1+(yx^{-1})^2)}$$
Therefore,
$$xfrac{partial z}{partial x}+yfrac{partial z}{partial y}=0$$

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
$$xfrac{partial z}{partial x}=frac{x}{1+left(frac{x}{y}right)^2}left(-frac{y}{x^2}right)=frac{1}{1+left(frac{x}{y}right)^2}left(-frac{y}{x}right)......(1)$$
$$yfrac{partial z}{partial y}=frac{1}{1+left(frac{x}{y}right)^2}left(frac{y}{x}right)..................(2)$$
adding (1)+(2),
$$xfrac{partial z}{partial x}+yfrac{partial z}{partial y}=frac{1}{1+left(frac{x}{y}right)^2}left[left(-frac{y}{x}right)+left(frac{y}{x}right)right]=0$$