Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when
$y=\frac{e^{\operatorname{ex}} \operatorname{sex} x \log x}{\sqrt{1-2 x}}$
Let $y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{\sqrt{1-2 x}}$
$\Rightarrow y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{(1-2 x)^{\frac{1}{2}}}$
Take log both sides:
$\Rightarrow \log y=\log \left(\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{(1-2 x)^{\frac{1}{2}}}\right)$
$\Rightarrow \log y=\log e^{2 x}+\log \sec ^{x} x+\log \log x-\log (1-2 x)^{\frac{1}{2}}$
$\left\{\log (a b)=\log a+\log b ; \log \left(\frac{a}{b}\right)=\log a-\log b\right\}$
$\Rightarrow \log y=a x \log e+x \log \sec x+\log \log x-\frac{1}{2} \log (1-2 x)\left\{\log x^{a}=a \log x\right\}$
$\Rightarrow \log y=a x+x \log \sec x+\log \log x-\frac{1}{2} \log (1-2 x)\{\log e=1\}$
Differentiating with respect to $\mathrm{x}$ :
$\left\{\right.$ Using chain rule $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where a and $\mathrm{u}$ are any variables $\}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a \frac{d x}{d x}+\left\{x \frac{d(\log \sec x)}{d x}+\log \sec x \frac{d x}{d x}\right\}+\frac{1}{\log x} \frac{d(\log x)}{d x}$
$-\frac{1}{2(1-2 x)} \frac{d(1-2 x)}{d x}$
$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a+\left\{x \times \frac{1}{\sec x} \frac{d(\sec x)}{d x}+\log \sec x\right\}+\frac{1}{\log x} \times \frac{1}{x} \frac{d x}{d x}-\frac{1}{2(1-2 x)}(-2)$
$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}\left(\mathrm{u}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{nu}^{\mathrm{n}-1} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a+\left\{\frac{x}{\sec x}(\sec x \tan x)+\log \sec x\right\}+\frac{1}{x \log x}-\frac{(-2)}{2(1-2 x)}$
$\left\{\frac{d(\sec x)}{d x}=\sec x \tan x\right\}$