mjhjmtu

Find the gradient of

$$z=x^y$$

I understand how to get it with respect to $x$ since $y$ is treated as a constant. But when trying to solve it with respect to $y$, why is it incorrect to implicitly differentiate and use the product rule:

$$ln(z)=ycdot ln(x)$$

$$frac(z_y)z=Bigr(ycdot frac1xBigr)+(1cdot ln(x))$$

$$z_y=zBigr(fracyx+ln(x)Bigr)$$

$$z_y=x^yBigr(fracyx+ln(x)Bigr)$$

In single-variable calculus, a first application of implicit differentiation is typically to find the derivative of $x mapsto a^x$, where $a > 0$. The typical argument is
beginalign*
y = a^x
&implies log(y) = xlog(a) \
&implies frac1y y' = log(a) \
&implies y' = ylog(a) = a^x log(a).
endalign*
In your problem, when you differentiate with respect to $y$, you need to regard $x$ as a constant (you should also probably assume that $x > 0$). You can then apply the single-variable result to get
$$ z_y
= fracmathrmdmathrmdy x^y
= x^ylog(x).
$$

Note that $x $ is not a function of $y $. Hence $x_y=0$. So the correct one is:

$$ln(z)=ycdotln(x)$$

$$frac(z_y)z=1cdotln(x)$$

$$z_y=zln(x)$$

$$z_y=x^yln(x)$$

When take the derivative of $x^y$ with respect to $y$, are you treating $x$ as a constant or as a function of $y$?

If the former then (using your notation) you get $fracz_yz=1 times ln(x)$ and so $z_y = x^y ln(x)$

If the latter then you get $fracz_yz=y times frac1xtimes x_y+1 times ln(x)$ and so $z_y =y x^y-1 x_y+ x^y ln(x)$. Then you might also want to say $z_x = y x^y-1+ x^y ln(x) y_x$ when taking the derivative of $x^y$ with respect to $x$

When we're differentiating with respect to $y$, $x$ should be treated as a constant. The $frac yx$ term in there that comes from differentiating $x$ shouldn't be there.

The variables $x$ and $y$ are to be considered independent when computing the gradient of $z$, that is,

$nabla z = (z_x, z_y), tag 1$

which is a vector. We have

$z = x^y, tag 2$

whence

$z_x = yx^y - 1, tag 3$

as our OP Random Student has noted. As for $z_y$, we have

$ln z = y ln x, tag 4$

whence

$dfracz_yz = ln x, tag 5$

or

$z_y = z ln x = x^y ln x; tag 6$

thus,

$nabla z = (z_x, z_y) = (yx^y - 1, x^y ln x). tag 7$

In (4)-(6), we have implicitly differentated $z$ with respect to $y$ alone, holding $x$ constant.