Differentiate w.r.t. x the function in Exercise : x^x+x^a+a^x+a^a… - Vidyadip

Question

Differentiate $\text{w.r.t. x}$ the function in Exercise :
$\text{x}^\text{x}+\text{x}^\text{a}+\text{a}^\text{x}+\text{a}^\text{a},$ for some fixed $a > 0$ and $x > 0$

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Answer

Let $\text{y}=\text{x}^\text{x}+\text{x}^\text{a}+\text{a}^\text{x}+\text{a}^\text{a}$
Also, let $x^x = u, x^a = v, a^x = w,$ and $a^a = s$
$\therefore y = u + v + w + s$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}+\frac{\text{ds}}{\text{dx}}$
$U = x^x$
$\Rightarrow\ \log\text{u}=\log\text{x}^\text{x}$
$\Rightarrow\ \log\text{u}=\text{x}\log\text{x}$
Differentiating both sides with respect to $x,$ we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}.\frac{\text{d}}{\text{dx}}(\text{x)}+\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\log\text{x}.1+\text{x}.\frac{1}{\text{x}}\Big]$
$V = x^a$
$\therefore\ \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{a})$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{ax}^{\text{a}-1}\ \dots(3)$
$W = a^x$
$\Rightarrow\ \log\text{w}=\log\text{a}^\text{x}$
$\Rightarrow\ \log\text{w}=\text{x}\log\text{a}$
Differentiating both sides with respect to $x,$ we obtain
$\Rightarrow\ \frac{\text{dw}}{\text{dx}}=\text{w}\log\text{a}$
$\Rightarrow\ \frac{\text{dw}}{\text{dx}}=\text{a}^\text{x}\log\text{a}\ \dots(4)$
$S = a^a$
Since a is constant$, a^a$ is also a constant.
$\therefore\ \frac{\text{ds}}{\text{dx}}=0\ \dots(5)$
From $(1), (2), (3), (4),$ and $(5),$ we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})+\text{ax}^{\text{a}-1}+\text{a}^\text{x}\log\text{a}+0$

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