Find derivative of
[tex]y = log_{2 } \: ( log_{2} \: x)[/tex]
by chain rule.
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Find derivative of
[tex]y = log_{2 } \: ( log_{2} \: x)[/tex]
by chain rule.
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Step-by-step explanation:
[tex]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[/tex]
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given function is
[tex]\red{\rm :\longmapsto\:y = log_{2}( log_{2}(x) )}[/tex]
On differentiating both sides w. r. t. x, we get
[tex]\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} log_{2}( log_{2}(x) )[/tex]
We know,
[tex] \boxed{ \bf{ \: \dfrac{d}{dx} log_{a}(x) = \frac{1}{x \: loga} }}[/tex]
So, using this result, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ log_{2}(x)log2 }\dfrac{d}{dx} log_{2}(x) [/tex]
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ log_{2}(x)log2 } \times \dfrac{1}{x \: log2} [/tex]
[tex]\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{ {(log2)}^{2} \: log_{2}(x) \: x} [/tex]
Additional Information :-
[tex]\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}[/tex]