If ϕ (x) = _ 8 _ 3 x, then ϕ ^ ′ (e) is equal to

We have,
ϕ \left( x\right) = \log_{8} \log_{3} x = \log_{8} \left(\dfrac{log x}{\log 3}\right)
$= \log_{8} \left(log x\right) - \log_{8} \left(\log 3\right)$
$= \dfrac{\log \left(log x\right)} {\log 8} - \log_{8}$ \left(\log 3\right)$<br />${ϕ}^{′} \left(x\right) = \dfrac{1}{\log 8}$⋅$\dfrac{1}{\log x}$⋅$\dfrac{1}{x} -0$<br />$∴ {ϕ}^{′} \left(e\right) = \dfrac{1}{\log 8}$⋅$\dfrac{1}{loge}$⋅$\dfrac{1}{e} = \dfrac{1}{elog8}