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If  y=x^{\log \left(x\right)^{\log \left(\log \left(x\right)\right)}} then \frac{dy}{dx} is equal to :

Option: 1

\frac{ylny}{xlnx}(2ln\: ln\: x+1)


Option: 2

\frac{xlnx}{ylny}(2ln\: ln\: x+1)


Option: 3

\frac{2ylny}{xlnx}(ln\: ln\: x+1)


Option: 4

None of these


Answers (1)

best_answer

 

Logarithmic functions -

\frac{d}{dx}(log_{e}x)=\frac{1}{x}


\frac{d}{dx}(log_{e}f(x))=\frac{1}{f(x)}.\frac{d}{dx}f(x)

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y=x^{\log \left(x\right)^{\log \left(\log \left(x\right)\right)}}

Taking \log_{e} both sides, We get lny= \left ( ln\: x \right )^{ln\left ( lnx \right )}.lnx

again taking \log_{e}, We get ln (ln y) = ln (ln x) . ln (ln x) + ln (ln x) =\left \{ {ln (ln x)} \right \} [ln (ln x) + 1]

Diff w.r.t x,

\left ( \frac{1}{ln\: y} \right )\left ( \frac{1}{y}.\frac{dy}{dx} \right )= \left ( \frac{1}{xlnx} \right )\left [ ln\left ( ln\: x \right )+1 \right ]+\left \{ ln(ln\: x) \right \}\left [ \frac{1}{xlnx} \right ]

\frac{dy}{dx}=y\: lny\left [ \frac{ln(lnx)}{xlnx} +\frac{1}{xlnx}+\frac{ln(lnx)}{xlnx}\right ]

\frac{dy}{dx}=y\: lny\left [ \frac{2ln(lnx)+1}{xlnx} \right ]

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manish

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