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Mathematics Question on Continuity and differentiability

If y=log⁡log⁡(log⁡ x)2y = \log \\{ \log ( \log \:x)^2 \\} then dydx\frac{dy}{dx} is equal to :

A

1log⁡(log⁡ x) \frac{1}{\log (\log \: x) }

B

1x log⁡(log⁡ x) \frac{1}{x \: \log (\log \: x) }

C

1x log⁡ x.log⁡(log⁡ x) \frac{1}{x \:\log \: x . \log (\log \: x) }

D

12x log⁡ 2x.log⁡(log⁡ x) \frac{1}{2x \:\log \: 2x . \log (\log \: x) }

Answer

1x log⁡ x.log⁡(log⁡ x) \frac{1}{x \:\log \: x . \log (\log \: x) }

Explanation

Solution

Let y=log⁡(log⁡(log⁡x)2)y = \log\left(\log \left(\log x\right)^{2}\right) \therefore \:\:\: \frac{dy}{dx}=\frac{d}{dx} \left[\log \left\\{\log \left(\log x\right)^{2}\right\\}\right] =1log⁡(log⁡x)2ddx[log⁡(log⁡x)2] = \frac{1}{\log \left(\log x\right)^{2}} \frac{d}{dx }\left[\log \left(\log x\right)^{2}\right] =1log⁡(log⁡x)2×1(log⁡x)2×ddx[(log⁡x)2]=\frac{ 1}{\log\left(\log x\right)^{2}} \times\frac{1}{\left(\log x\right)^{2}} \times\frac{d}{dx}\left[\left(\log x\right)^{2}\right] =1log⁡(log⁡x)2×1(log⁡x)2×2log⁡xx = \frac{1}{\log \left(\log x\right)^{2}} \times\frac{1}{\left(\log x\right)^{2}} \times\frac{2 \log x}{x} =2xlog⁡x.log⁡(log⁡x)2= \frac{2}{x \log x.\log \left(\log x\right)^{2}} =22xlog⁡xlog⁡(log⁡x)=1xlog⁡xlog⁡(log⁡x)= \frac{2}{2x \log x \log \left(\log x\right)} = \frac{1}{x \log x \log \left(\log x\right)}