Question

3. Differentiate the following functions w.r.t. x :
(i) elog(log x) . log 3x
Solution :
elog) log 3x = log x. log 3x|
•alogat​

Answer 1

SOLUTION

TO DETERMINE

To Differentiate

$\displaystyle \sf{ {e}^{ log( log x) }. log 3x }$

EVALUATION

Let

$\displaystyle \sf{y = {e}^{ log( log x) }. log 3x }$

$\displaystyle \sf{ \implies \: y = log x \: . \: log 3x }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = log x \: . \frac{d}{dx} \bigg( log 3x \bigg) + log 3x \: . \frac{d}{dx} \bigg( log x \bigg) }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = log x \: . \frac{3}{3x} + log 3x \: . \frac{1}{x} }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = log x \: . \frac{1}{x} + log 3x \: . \frac{1}{x} }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{1}{x} \bigg( log x + log 3x \bigg) }$

FINAL ANSWER

$\displaystyle \sf{ \frac{dy}{dx} = \frac{1}{x} \bigg( log x + log 3x \bigg) }$

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