Derivative of $y=x^y$

Question

How to find the function that gives the slope of the line tangent to $y = x^y$, at a certain $x$ value?

Unless I made a mistake:

$$\frac{dy}{dx}=\frac{ye^{y\ln x}}{\ln x-\ln^2x}$$

How would I use that to graph the function that gives the slope of the tangent line along $y = x^y$ for any given $x$?

Thanks

Answer 1

1. I think you made a mistake: $$\begin{align*} y&= x^y\\ y &= e^{y\ln x}\\ y’ &= (y\ln x)’e^{y\ln x}\\ y’ &= \left(y’\ln x + \frac{y}{x}\right) x^y\\ y’ &= y’(x^y\ln x) + \frac{yx^y}{x}\\ y’ - y’(x^y\ln x) &= yx^{y-1}\\ y’(1 - x^y\ln x) &= yx^{y-1}\\ y’ &= \frac{yx^{y-1}}{1-x^y\ln(x)}. \end{align*}$$

(To see this is the same as the other answer, replace the $y$ in the numerator with $x^y$)

2. Given a value of $x$, you use the equation $y=x^y$ to find the corresponding value of $y$ (this is difficult sometimes for this function). That will give you the point. Then plug in the values of $x$ and $y$ into the equation to get the slope. Exactly the same way you do with any implicit differentiation. For example, if $x=1$, you want to solve $y=1^y$; the obvious answer is $y=1$. Plugging into the formula, we get $$y’\Bigm|_{(1,1)} = \frac{1(1^0)}{1 - (1^1)\ln(1)} = \frac{1}{1} = 1,$$ so you want the line with slope $1$ that goes through $(1,1)$.

Answer 2

hint: $y=x^y$ implies $x=y^{1/y}$, you wouldn’t need implicit differentiation if you considered $x$ as a function of $y$.

also you can use the logarithm to differentiate implicitly $\frac{d}{dx}[y]=\frac{d}{dx}[x^y]=\frac{d}{dx}[e^{y\cdot \ln(x)}]=e^{y\cdot\ln(x)}\cdot \frac{d}{dx}[y\cdot\ln(x)]=x^y\left[ \frac{d}{dx}[y]\cdot \ln(x)+\frac{x^y}{x}\right]$

which after solving for $\frac{d}{dx}[y] = \frac{x^{2y-1}}{1-x^y\cdot\ln(x)}$ would give you the slope for every point $(x,x^y)$ on the curve, so now instead of having to plug in only $x$ to find the slope of the curve, you would have to plug in the appropriately calculated value of $y$,

you could probably do that more easily by first choosing a $y$, then solving for $x$, so it’s much easier to know the slope at a certain $y$ value than an $x$ value in this case.