Question

Find derivative of
$y = log_{2 } \: ( log_{2} \: x)$
by chain rule.​

Answer 1

Step-by-step explanation:

$y = log_{2}( log_{2}(x) ) \\ = log_{2}( \frac{ log_{e}(x) }{ log_{e}(2) } ) \\ = log_{2}( log_{e}(x) ) - log_{2}( log_{e}(2) ) \\ = \frac{ log_{e}( log_{e}(x) ) }{ log_{e}(2) } - \frac{ log_{e}( log_{e}(2) ) }{ log_{e}(2) } \\ = \frac{1}{ log_{e}(2) } \frac{d \: \: log_{e}( log_{e}(x) ) }{dx} - 0 \\ = \frac{1}{ log_{e}(2) } \frac{1}{ log_{e}(x) \times x} - 0$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\red{\rm :\longmapsto\:y = log_{2}( log_{2}(x) )}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} log_{2}( log_{2}(x) )$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx} log_{a}(x) = \frac{1}{x \: loga} }}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ log_{2}(x)log2 }\dfrac{d}{dx} log_{2}(x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ log_{2}(x)log2 } \times \dfrac{1}{x \: log2}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{ {(log2)}^{2} \: log_{2}(x) \: x}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$