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Here, we have to differentiate the equation $y={{\log }_{7}}\left( \log x \right)$ with respect to x.

But we are having the equation in base 7. So, first we will apply the change base rule to convert base 7 into base ‘e’ i.e. natural base using the formula ${{\log }_{a}}b=\dfrac{{{\log }_{n}}b}{{{\log }_{n}}a}$

Here, we have $b=\log x$ and a as 7.

So, on using the rule, we can write it as

$y={{\log }_{7}}\left( \log x \right)=\dfrac{{{\log }_{e}}\left( \log x \right)}{{{\log }_{e}}7}$ …………………..(1)

Now, we will differentiate both sides with respect to x. So, we get as

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{{{\log }_{e}}\left( \log x \right)}{{{\log }_{e}}7} \right)$

Here, ${{\log }_{e}}7$ is constant term so we will take it out of differentiation and we know that $\dfrac{d}{dx}{{\log }_{e}}x=\dfrac{1}{x}$ . So, using this formula we will get equation as

$\dfrac{dy}{dx}=\dfrac{1}{{{\log }_{e}}7}\dfrac{d}{dx}{{\log }_{e}}\left( \log x \right)$

Now, we will be applying chain rule in differentiating $\dfrac{d}{dx}{{\log }_{e}}\left( \log x \right)$ . So, we get as

$\dfrac{dy}{dx}=\dfrac{1}{{{\log }_{e}}7}\times \dfrac{1}{\log x}\times \dfrac{d}{dx}\left( \log x \right)$

We also know that $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$ . On putting values, we get as

$\dfrac{dy}{dx}=\dfrac{1}{{{\log }_{e}}7}\times \dfrac{1}{\log x}\times \dfrac{1}{x}$

Thus, final answer will be

$\dfrac{dy}{dx}=\dfrac{1}{x{{\log }_{e}}7\left( \log x \right)}$

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