Whats the derivative of y=In(In(In(Inx)))

**tenner** · April 10th 2012, 04:46 PM

I think it is y'= (1)/(x)(In(In(Inx))) but I'm not sure.

**BAdhi** · April 10th 2012, 08:00 PM

no. From the chain rule, you should get,

$y'=\frac{1}{\ln(\ln(\ln x))}.\frac{1}{\ln(\ln x)}.\frac{1}{\ln x}.\frac{1}{x}$

**Prove It** · April 10th 2012, 08:01 PM

Originally Posted by tenner

I think it is y'= (1)/(x)(In(In(Inx))) but I'm not sure.

First of all, the function is not "In", it's "ln", which stands for "natural logarithm". Anyway

$\displaystyle \begin{align*} y &= \ln{\left\{\ln{\left[\ln{\left(\ln{x}\right)}\right]}\right\}} \end{align*}$

Let $\displaystyle \begin{align*} u = \ln{x} \implies y = \ln{\left[\ln{\left(\ln{u}\right)}\right]} \end{align*}$

Let $\displaystyle \begin{align*} v = \ln{u} \implies y = \ln{\left(\ln{v}\right)} \end{align*}$

Let $\displaystyle \begin{align*} w = \ln{v} \implies y = \ln{w} \end{align*}$

Then

$\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{du}{dx}\cdot \frac{dv}{du} \cdot \frac{dw}{dv} \cdot \frac{dy}{dw} \\ &= \frac{1}{x} \cdot \frac{1}{u} \cdot \frac{1}{v} \cdot \frac{1}{w} \\ &= \frac{1}{x} \cdot \frac{1}{\ln{x}} \cdot \frac{1}{\ln{u}} \cdot \frac{1}{\ln{v}} \\ &= \frac{1}{x} \cdot \frac{1}{\ln{x}} \cdot \frac{1}{\ln{\left(\ln{x}\right)}} \cdot \frac{1}{\ln{\left(\ln{u}\right)}} \\ &= \frac{1}{x} \cdot \frac{1}{\ln{x}} \cdot \frac{1}{\ln{\left(\ln{x}\right)}} \cdot \frac{1}{\ln{\left[\ln{\left(\ln{x}\right)}\right]}} \end{align*}$

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