What is the derivative of $x^{(x^2)}$?

Question

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.

Substitute $x^2$=u and then do implicit differentiation ie $dy/du.du/dx$ — Archis Welankar, Nov 08 '15 at 05:06

cheesyfluff · Answer 1 · 2015-11-08T04:52:33.643

4

\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}

edited Nov 08 '15 at 04:52

answered Nov 08 '15 at 04:42

cheesyfluff

1,745
11
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score 4 · Accepted Answer · answered Nov 08 '15 at 04:43

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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?

answered Nov 08 '15 at 04:43

graydad

14,077

score 1 · Answer 3 · answered Nov 08 '15 at 04:44

1

$$y=x^{(x^{2})}$$

$$ln(y)=x^{2}ln(x)$$

$$\frac{1}{y}\frac{dy}{dx}=2xln(x)+x^{2}\frac{1}{x}$$

$$\frac{dy}{dx}=x^{(x^{2})}(2xln(x)+x)$$

answered Nov 08 '15 at 04:44

Jerry Guern

2,662

score 0 · Answer 4 · answered Nov 08 '15 at 05:34

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$% Predefined Typography % \newcommand{\iint}[2]{\int {#1}~{\rm d}{#2}} \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:

$$\begin{align} \pred{g(x, f(x))}{x} &= f(x)~x^{f(x)-1} + \ln(x)~x^{f(x)} \cdot 2x \\ &= x^2~x^{x^2-1} + \ln(x)~x^{x^2} \cdot 2x \\ &= x~x^{x^2} + \ln(x)~x^{x^2} \cdot 2x \\ \\ &= x^{x^2}\bigg(x + 2x~\ln(x)\bigg) \end{align}$$

answered Nov 08 '15 at 05:34

DanielV

23,556

A neat trick, but almost certainly beyond the curriculum of someone learning derivatives for the first time – Ben Grossmann Nov 22 '15 at 13:38
Comparing all the answers, this is the one you would call a "trick"? The most algorithmic and complete way to compute a derivative expression? I agree the notation is not taught in first courses, but the concept of chain rule is and it is just a notational difference. If I asked how to compute an expression, and someone replied with a only-works-this-time response (use $\ln$) rather than a general approach (chain rule), I'd be quite annoyed. I am very sympathetic to the problem of users posting "beyond the cirriculum" responses though, and will keep it in mind. – DanielV Nov 22 '15 at 16:43
I'd call any approach here a "trick", I didn't mean that as a barb. It's a nice approach for those who know the different forms of the chain rule. The difference between chain rules, however, is far more than a "symbolic" difference, I would argue. It is not clear how one should justify simply taking the different derivatives and adding them together. There is a subtle something under the hood there. – Ben Grossmann Nov 22 '15 at 23:28

score 0 · Answer 5 · answered Nov 22 '15 at 13:16

By definition, $x^{(x^2)}=e^{x^2\log x} $. Now we differentiate, using the chain rule: $$ \frac {d (x^{(x^2)})}{dx}=\frac {d (e^{x^2\log x} )}{dx} =\frac {d ({x^2\log x} )}{dx}\,e^{x^2\log x} =\left(2x\log x+\frac {x^2}{x}\right)\,e^{x^2\log x} =\left(2x\log x+{x}\right)\,e^{x^2\log x} =\left(2x\log x+{x}\right)\,x^{x^2} . $$

What is the derivative of $x^{(x^2)}$?

5 Answers5