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Question

Differentiate the given function w.r.t. $$x$$.
$$\displaystyle \log(\log x) , x > 1$$


Solution

Let $$\displaystyle y = \log(\log x) $$
Thus using chain rule,
$$\displaystyle \frac{dy}{dx} = \frac{d}{dx} [ \log(\log x) ]$$
$$\displaystyle = \frac{1}{\log x} . \frac{d}{dx} (\log x) $$
$$\displaystyle = \frac{1}{\log x} . \frac{1}{x}$$
$$\displaystyle = \frac{1}{x \log x} , x > 1$$

Mathematics
RS Agarwal
Standard XII

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