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Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]4 Marks
Question
Find $\frac{ d y}{ d x}$, if $y =(\log x )^{ x }+( x ) \log ^{ x }$
✓

Answer

$y=(\log x)^x+(x) \log x$
Let $u =(\log x )^{ x }$ and $v = x \log x$
$\therefore y = u + v$
Differentiating both sides w. r. t. x, we get
$\frac{ d y}{ d x}=\frac{ du }{ d x}+\frac{ dv }{ d x} \ldots(\text{i})$
Now, $u=(\log x)^x$
Taking logarithm of both sides, we get
$\log u=\log (\log x)^x=x \log (\log x)$
Differentiating both sides w. r. t. x, we get
$ \frac{ d }{ d x}(\log u )=x \cdot \frac{ d }{ d x}[\log (\log x)]+\log (\log x) \cdot \frac{ d }{ d x}(x)$
$\therefore \frac{1}{ u } \cdot \frac{ du }{ d x}=x \cdot \frac{1}{\log x} \cdot \frac{ d }{ d x}(\log x)+\log (\log x) \cdot 1 $
$ \therefore \frac{1}{ u } \cdot \frac{ du }{ d x}=x \cdot \frac{1}{\log x} \cdot \frac{1}{x}+\log (\log x)$
$\therefore \frac{ du }{ d x}= u \left[\frac{1}{\log x}+\log (\log x)\right]$
$\therefore \frac{ du }{ d x}=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right] \ldots(\text{ii}) $
Also, $v=x^{\log x}$
Taking logarithm of both sides, we get
$ \log v=\log \left(x^{\log x}\right)=\log x(\log x)$
$\therefore \log v=(\log x)^2 $
Differentiating both sides w.r.t. $x$, we get
$ \frac{1}{ v } \cdot \frac{ dv }{ d x}=2 \log x \cdot \frac{ d }{ d x}(\log x)$
$\therefore \frac{1}{ v } \cdot \frac{ dv }{ d x}=2 \log x \cdot \frac{1}{x} $
Substituting (ii) and (iii) in (i), we get.: $\frac{ d }{ d x}= v \left[\frac{2 \log x}{x}\right]$
$\therefore \frac{ dv }{ d x}=x^{\log x}\left[\frac{2 \log x}{x}\right]$
Substituting (ii) and (iii) in (i), we get
$\frac{ d y}{ d x}=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]+x^{\log x}\left[\frac{2 \log x}{x}\right]$

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