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

Let y=x^x+x^a+a^x+a^aAlso, let x^x = u, x^a = v, a^x = w, and a^a = s y = u + v + w + s dy dx = du dx + dv dx + dw dx + ds dx U = x^x = ^x =x Differentiating both sides with respect to x, we obtain 1 u du dx = . d dx (x)+x. d dx ( ) du dx =u [ .1+x.1 x ] V = x^a dv dx = d dx (x^a) dv dx =ax^ a-1 (3) W = a^x = ^x =x Differentiating both sides with respect to x, we obtain dw dx =w dw dx =a^x (4) S = a^a Since a is constant, a^a is also a constant. ds dx =0 (5) From (1), (2), (3), (4), and (5), we obtain dy dx =x^x(1+ )+ax^ a-1 +a^x +0

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

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Question

Differentiate 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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