Derivative of $ 4e^{xy ^ {y}} $

Question

Here's what I've got so far: for $ y $ as some function of x:

$$ f(x) =4e^{(xy) ^ {y}} \\ f'(x) = 4e^{(xy) ^ {y}} \cdot \frac {d}{dx} [(xy)^{y}] \cdot\frac{d}{dx} [y]\\ \implies 4e^{(xy) ^ {y}} \left(y + x\frac{dy}{dx}\right) \cdot \frac{dy}{dx}. $$

Answer 1

\begin{align} f&=4e^{(xy)^y}\\ \implies f'&=\frac{d}{dx}\left[4e^{{(xy)}^y}\right]\\ &=4e^{{(xy)}^y}\cdot\frac{d}{dx}(xy)^y\\ &=4e^{{(xy)}^y}\cdot\frac{d}{dx}\left[e^{y\ln(xy)}\right]\\ &=4e^{{(xy)}^y}\cdot e^{y\ln(xy)}\cdot\frac{d}{dx}\left[y\ln(xy)\right]\\ &=4e^{{(xy)}^y}\cdot(xy)^y\cdot\left[\frac{dy}{dx}\cdot\ln(xy)+\frac{y}{xy}\cdot\frac{d}{dx}(xy)\right]\\ &=4e^{{(xy)}^y}\cdot(xy)^y\cdot\left[\frac{dy}{dx}\cdot\ln(xy)+\frac1x\cdot\left(y+x\cdot\frac{dy}{dx}\right)\right]\\ \implies f'&=4e^{{(xy)}^y}\cdot(xy)^y\cdot\left[\frac{dy}{dx}\cdot\ln(xy)+\frac yx+\frac{dy}{dx}\right]\\ \end{align}

Answer 2

First, we need to figure out the derivative of $h(x)=x^x$. We have $$\ln h = x\ln x$$ $$\frac {h'}h = 1+\ln x$$ $$\implies h' = (1+\ln x)x^x$$ Using the chain rule, it follows that $(y^y)' = y'y^y(1+\ln y)$. Further, noting that for any $g(x)$ the derivative of $e^g$ is $g'e^g$, we have $$\frac{d}{dx}x^y = \frac{d}{dx}e^{y\ln x} = \left(\frac y x+y'\ln x\right)x^y$$ Finally,

$$\frac{d}{dx}\left(4e^{(xy)^y}\right) = 4 \left(\frac{d}{dx}x^yy^y\right)e^{(xy)^y}$$ $$ = 4\left(\left(\frac y x+y'\ln x\right)x^yy^y+x^yy^yy'(1+\ln y)\right)e^{(xy)^y}$$ $$= 4\left(\frac y x+y'(1+\ln x+\ln y) \right)(xy)^ye^{(xy)^y}$$

Answer 3

I like to define simple functions and define the $f(x)$ as a composition of these simple functions. Then I don't get lost when applying the chain rule.

let $u = xy\\ v = u^y = e^{y\ln u}$

$f(x) = 4e^v\\ f' = 4e^v v'\\ v' = u^y(y'\ln u + \frac {y}{u}u')\\ u' = y + xy'$

$f'= 4e^{(xy)^y}(xy)^y(y'(\ln x + \ln y) + \frac {y}{x}+ y')$