What is the derivative of $x^{(x^2)}$? What is the derivative of $x^{(x^2)}$? I'm having difficulty with this question because I keep computing $y'=e^{xlnx}e^{2lnx}$ but the I graph it: https://www.desmos.com/calculator/u5vm44kedt and it doesn't look right.
 A: \begin{gather}
y=e^{x^2\ln x}\\
y^\prime = e^{x^2\ln x} \frac{d}{dx} (x^2 \ln x)
=e^{x^2\ln x}(2x\ln x + x)\\
=x^{(x^2)}(2x\ln x + x)
\end{gather}
A: Differentiate $$\ln(y) = x^2\ln(x)$$ implicitly to get a result. You should get $$\frac{y'}{y} = 2x\ln(x) + x$$ before eliminating $y$. Can you continue from here?
A: $$y=x^{(x^{2})}$$
$$ln(y)=x^{2}ln(x)$$
$$\frac{1}{y}\frac{dy}{dx}=2xln(x)+x^{2}\frac{1}{x}$$
$$\frac{dy}{dx}=x^{(x^{2})}(2xln(x)+x)$$
A: $% Predefined Typography %
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 \newcommand{\dint}[4]{\int_{#3}^{#4}{#1}~{\rm d}{#2}}
 \newcommand{\pred}[2]{\frac{\rm d}{{\rm d}{#2}}#1}
 \newcommand{\ind} [2]{\frac{{\rm d} {#1}}{{\rm d}{#2}}}
 \newcommand{\predp}[2]{\frac{\partial}{\partial {#2}}#1}
 \newcommand{\indp} [2]{\frac{{\partial} {#1}}{\partial {#2}}}
 \newcommand{\predn}[3]{\frac{\rm d}^{#3}{{\rm d}{#2}^{#3}}#1}
 \newcommand{\indn} [3]{\frac{{\rm d}^{#3} {#1}}{{\rm d}{#2}^{#3}}}
$
Let $g(y, z) = y^z$.
Let $f(w) = w^2$.
You want to compute $\pred{g(x, f(x))}{x}$.
Use the chain rule to get
$$\pred{g(x, f(x))}{x} = \indp{g}{y}(x, f(x)) + \indp{g}{z}(x, f(x)) \cdot \indp{f}{w}(x)$$
We have that 
$$\indp{g}{y}(y, z) = z~y^{z-1}$$
$$\indp{g}{z}(y, z) = \ln(y)~y^z$$
$$\indp{f}{w}(w) = 2w$$
So plugging those in you get:
$$\begin{align}
\pred{g(x, f(x))}{x} &= f(x)~x^{f(x)-1}  + \ln(x)~x^{f(x)} \cdot 2x \\
&= x^2~x^{x^2-1}  + \ln(x)~x^{x^2} \cdot 2x \\
&= x~x^{x^2}  + \ln(x)~x^{x^2} \cdot 2x \\ \\
&= x^{x^2}\bigg(x  + 2x~\ln(x)\bigg)
\end{align}$$
A: By definition, $x^{(x^2)}=e^{x^2\log x} $. Now we differentiate, using the chain rule:
$$
\frac {d (x^{(x^2)})}{dx}=\frac {d (e^{x^2\log x} )}{dx}
=\frac {d ({x^2\log x} )}{dx}\,e^{x^2\log x} 
=\left(2x\log x+\frac {x^2}{x}\right)\,e^{x^2\log x} 
=\left(2x\log x+{x}\right)\,e^{x^2\log x} 
=\left(2x\log x+{x}\right)\,x^{x^2} .
$$
