Question

solve fastttt it ??!!!!???!! ​

Answer 1

GIVEN :–

• A function $\to \bf log_{ex}{logx}$

TO FIND :–

• Differentiate form with respect to 'x' = ?

SOLUTION :–

• Let the function –

$\\ \implies \bf y = log_{ex}{logx}$

• Using identity –

$\\ \implies \bf log_{a}{b} = \dfrac{ log_{e}(b) }{ log_{e}(a) }$

• So –

$\\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(ex)}$

• Using identity –

$\\ \implies \bf log_{e}(a.b) = log_{e}(a)+log_{e}(b)$

• So –

$\\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(e) +log_{e}(x)}$

$\\ \implies \bf y = \dfrac{log_{e}(logx)}{ 1 +log_{e}(x)}$

• Now Differentiate with respect to 'x' –

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{[1 +log_{e}(x)] \bigg \{\dfrac{1}{x log_{e}(x)} \bigg \}- [log_{e}(logx)] \bigg( \dfrac{1}{x} \bigg) }{ [1 +log_{e}(x)]^{2} }$

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{\bigg \{\dfrac{1}{x log_{e}(x)}+ \dfrac{1}{x} \bigg\}- \dfrac{log_{e}(logx)}{x}}{ [1 +log_{e}(x)]^{2} }$

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{\dfrac{1}{x} \bigg \{\dfrac{1}{ log_{e}(x)}+1 - log_{e}(logx)\bigg\}}{ [1 +log_{e}(x)]^{2} }$

Answer 2

Answer:

SOLUTION :–

• Let the function –

$\\ \implies \bf y = log_{ex}{logx}$

• Using identity –

$\\ \implies \bf log_{a}{b} = \dfrac{ log_{e}(b) }{ log_{e}(a) }$

• So –

$\\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(ex)}$

• Using identity –

$\\ \implies \bf log_{e}(a.b) = log_{e}(a)+log_{e}(b)$

• So –

$\\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(e) +log_{e}(x)}$

$\\ \implies \bf y = \dfrac{log_{e}(logx)}{ 1 +log_{e}(x)}$

• Now Differentiate with respect to 'x' –

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{[1 +log_{e}(x)] \bigg \{\dfrac{1}{x log_{e}(x)} \bigg \}- [log_{e}(logx)] \bigg( \dfrac{1}{x} \bigg) }{ [1 +log_{e}(x)]^{2} }$

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{\bigg \{\dfrac{1}{x log_{e}(x)}+ \dfrac{1}{x} \bigg\}- \dfrac{log_{e}(logx)}{x}}{ [1 +log_{e}(x)]^{2} }$

$\\ \implies \bf \dfrac{dy}{dx} = \dfrac{\dfrac{1}{x} \bigg \{\dfrac{1}{ log_{e}(x)}+1 - log_{e}(logx)\bigg\}}{ [1 +log_{e}(x)]^{2} }$