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# If $y={{\log }_{7}}\left( \log x \right)$ then, find $\dfrac{dy}{dx}$ .  Hint: We will apply the change base rule to convert base 7 to base ‘e’ by using the formula ${{\log }_{a}}b=\dfrac{{{\log }_{n}}b}{{{\log }_{n}}a}$. On applying this formula, we will get equation like $y={{\log }_{7}}\left( \log x \right)=\dfrac{{{\log }_{e}}\left( \log x \right)}{{{\log }_{e}}7}$. Then, we will differentiate both sides with respect to x and on solving we will get the answer. Formula used here are $\dfrac{d}{dx}{{\log }_{e}}x=\dfrac{1}{x}$, $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$.

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

Note: Be careful while applying the change base formula. Sometimes instead of using ${{\log }_{a}}b=\dfrac{{{\log }_{n}}b}{{{\log }_{n}}a}$ , students take ${{\log }_{a}}b=\dfrac{{{\log }_{n}}a}{{{\log }_{n}}b}$ and whole answer gets wrong. If we use this wrong formula, then there are chances of getting incorrect answers and students will get confused while solving the equation. So, please remember the formula properly and then try to solve it.
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