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1)$$\ (x^x)^x$$ 2)$$\ x^{(x)^x}$$

consider answering this please ,as I am confused how to use chain rule here.

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    Sorry if it comes as duplicate... – 2017-02-25
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    Rewrite using $a^b = e^{b \log a}$. Then you can use the chain rule and multiplication rule. – 2017-02-25
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    try to use $x^x = \exp^{x\ln x}$ – 2017-02-25
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    Funny how it asks to differentiate a function, yet is tagged (integration). I'll edit that for you. – 2017-02-25
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    @S.C.B. I was doing integration ,lol so that topic was in my mind.. – 2017-02-25
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    partial duplicate of (http://math.stackexchange.com/questions/1317314/derivative-of-xxx-to-infinity) and (https://www.purplemath.com/learning/viewtopic.php?t=188) – 2017-02-25

5 Answers 5

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HINT

Note that $$ (x^x)^x=e^{x^2\ln(x)} $$ and$$ x^{x^x}=e^{x^x\ln x} =e^{\ln(x)e^{x \ln x}} $$ Use the product rule, and the fact the derivative of $e^{f(x)}$ is $$f'(x)e^{f(x)}$$ The last follows from chain rule.

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Well (solving problem 1), we want to find:

$$\frac{\text{d}}{\text{d}x}\left[\left(x^x\right)^x\right]\tag1$$

Express $\left(x^x\right)^x$ as a power of $e$:

$$\left(x^x\right)^x=\exp\left(\ln\left[\left(x^x\right)^x\right]\right)=\exp\left(x\ln\left[x^x\right]\right)\tag2$$

Using the chain rule:

$$\frac{\text{d}}{\text{d}x}\left[\left(x^x\right)^x\right]=\frac{\text{d}}{\text{d}x}\left[\exp\left(x\ln\left[x^x\right]\right)\right]=\exp\left(x\ln\left[x^x\right]\right)\cdot\frac{\text{d}}{\text{d}x}\left(x\ln\left[x^x\right]\right)\tag3$$

Now, using the product rule:

$$x\cdot\frac{\text{d}}{\text{d}x}\left(\ln\left[x^x\right]\right)+\ln\left[x^x\right]\cdot\frac{\text{d}}{\text{d}x}\left(x\right)\tag4$$

The derivative of $x$ equals $1$ and we also get using the product rule:

$$\frac{\text{d}}{\text{d}x}\left(\ln\left[x^x\right]\right)=\frac{\text{d}}{\text{d}x}\left(x\ln\left[x\right]\right)=x\cdot\frac{\text{d}}{\text{d}x}\left(\ln\left[x\right]\right)+\ln\left[x\right]\cdot\frac{\text{d}}{\text{d}x}\left(x\right)\tag5$$

The derivative of $\ln\left(x\right)$ is $\frac{1}{x}$.

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Standard chain rule:

1) Note that $x=e^{\ln(x)}$ so that

$$(x^x)^x=x^{x^2}=e^{x^2\ln(x)}$$

We may now apply chain rule to get

$$\frac d{dx}x^2\ln(x)=2x\ln(x)+x$$

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

2)

For the second, we rewrite in a similar manner:

$$x^{x^x}=x^{e^{x\ln(x)}}=e^{\ln(x)e^{x\ln(x)}}$$

We may now apply chain rule:

$$\frac d{dx}x\ln(x)=\ln(x)+1$$

$$\frac d{dx}\ln(x)e^{x\ln(x)}=\left[\ln(x)(\ln(x)+1)+\frac1x\right]e^{x\ln(x)}$$

$$\frac{dy}{dx}=e^{\ln(x)e^{x\ln(x)}}\left[\ln(x)(\ln(x)+1)+\frac1x\right]e^{x\ln(x)}=x^{x^x}x^x\left(\ln(x)(\ln(x)+1)+\frac1x\right)$$

Multivariable chain rule:

1) Let $y=u^v$, $u=x$, and $v=x^2$.

$$\begin{align}\frac{dy}{dx}&=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}\\&=vu^{v-1}+\ln(u)u^v2x\\&=x^{x^2}(2x\ln(x)+x)\end{align}$$

2) Let $y=u^v$, $u=x$, $v=t^z$, $t=x$ and $z=x$.

$$\begin{align}\frac{dy}{dx}&=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}\\&=vu^{v-1}+\ln(u)u^v\left[\frac{\partial v}{\partial t}\frac{dt}{dx}+\frac{\partial v}{\partial z}\frac{dz}{dx}\right]\\&=x^{x^x}(x^{x-1}+\ln(x)x^x(\ln(x)+1))\end{align}$$

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  1. Let $y = x^{x^2}$

Taking log on both sides,

$\log y = x^2 \log x$

Now you can find derivative.

  1. Let $y = x^{x^x}$

Taking log on both sides,

$\log y = x^x \log x$

Taking log again on both sides,

$\log \log y = x \log x + \log \log x$

Now you can find derivative.

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The first question is simply $$x^{x^2}$$ So to differentiate this in one line $$2xx^{x^2}ln(x)+x^2x^{x^2-1}$$.In any differentiation of form $$f(x)^{g(x)}$$ use $$f(x)^{g(x)}ln(f(x))g'(x)+g(x)f(x)^{g(x)-1}g'(x)$$.You can use the same technique for the other question.Note this is a method only to get quicker answers