CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})+\text{ax}^{\text{a}-1}+\text{a}^\text{x}\log\text{a}+0$