Question

differentiate with respect to x : e^log[(log x)^2-log x^2]​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Solution:

Before solving this question let's simplify our question::

$e^{log x} \\Let \ y = e^{log x}\\\\Apply \ log \ to \ both \ sides\\log(y) = log e^{log x}\\log(y) = log(x) * log(e) \ We \ know, [log \ rule \ log e^{x} = x log e] and \ loge = 1\\log(y) = log(x) \ [Both \ have \ log \ which \ means \ they \ are \ equal]\\y = x$

This gets proved that $e^{logx} = x$

Now, lets solve our question,

$e^{log[log^{2}x - logx^{2} ] }$

After simplification it will be,

$[log^{2}x - logx^{2} ]$

Now, let's apply dy/dx (i.e. differentiate with respect to x) on it.

$\frac{d}{dx} [log^{2}x - logx^{2} ]$

$\frac{d}{dx} [log^{2}x ] -2 [\frac{d}{dx} logx]$

$2 log (x) \frac{d}{dx} [log(x)] - 2(\frac{1}{x})$

$2 log (x) * \frac{1}{x} - \frac{2}{x}$

$\frac{2log(x)}{x} - \frac{2}{x}$

$\frac{2 log(x) - 2}{x}$

The differentiation of $e^{log[log^{2}x - logx^{2} ] }$ will be $\frac{2 log(x) - 2}{x}$