# Why is my logic incorrect? I think it is right, my answer is ALMOST the same. Derivatives and chain rule.

I want to find the derivative of $y = \log_b(\log_b(x))$

I am going to let $u = \log_b(x)$ so that $y = \log_b(u)$. BY Chain Rule, I get

$$\frac{dy}{dx} =\frac{dy}{du} \frac{du}{dx}$$

$$\frac{dy}{du} = \frac{1}{u\ln(b)} = \frac{1}{\log_b(x)\ln(b)}$$

$$\frac{du}{dx} = \frac{1}{x\ln(b)}$$

$$\frac{dy}{dx} =\frac{dy}{du} \frac{du}{dx} =\frac{1}{\log_b(x)\ln(b)} \frac{1}{x\ln(b)} = \frac{1}{\ln^2(b)x\log_b(x)}$$

The answer I got from Wolframalpha is http://www.wolframalpha.com/input/?i=D[log[log%28x%29]%2Cx]&a=*C.D-_*Function.dflt-&a=*FunClash.log-_*Log10.Log-

Which they only have one $\ln(b)$.

Note that they have $b = 10$ in this case.

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No, I have $ln^2 (10)$ –  mark Mar 2 '12 at 2:41
Just keep using the chain rule as necessary. For example $${d\over dx}\color{maroon}\ln\bigl(\color{darkgreen}\ln(\color{orange}\ln (x^2))\bigr) ={1\over\color{darkgreen}\ln(\color{orange}\ln(x^2))}\cdot{1\over \color{orange}\ln(x^2)}\cdot{1\over x^2}\cdot2x.$$ –  David Mitra Mar 2 '12 at 3:19

Ahh, I got it. $$\frac 1{x \log(b)^2 \log_b(x)} = \frac 1{x \log(b) \log(x)}$$ because $\log_b (x) = \log(x) / \log(b)$. Therefore one $\log(b)$ goes away there.