Solve this


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, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a+\left\{x \times \frac{1}{\sec x} \frac{d(\sec x)}{d x}+\log \sec x\right\}+\frac{1}{\log x} \times \frac{1}{x} \frac{d x}{d x}-\frac{1}{2(1-2 x)}(-2)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}\left(\mathrm{u}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{nu}^{\mathrm{n}-1} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a+\left\{\frac{x}{\sec x}(\sec x \tan x)+\log \sec x\right\}+\frac{1}{x \log x}-\frac{(-2)}{2(1-2 x)}$

$\left\{\frac{d(\sec x)}{d x}=\sec x \tan x\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a+\left\{\frac{x \sec x \tan x}{\sec x}+\log \sec x\right\}+\frac{1}{x \log x}+\frac{1}{(1-2 x)}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a+\{x \tan x+\log \sec x\}+\frac{1}{x \log x}+\frac{1}{(1-2 x)}$

$\Rightarrow \frac{d y}{d x}=y\left\{a+x \tan x+\log \sec x+\frac{1}{x \log x}+\frac{1}{(1-2 x)}\right\}$

Put the value of $y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{\sqrt{1-2 x}}$ :

$\Rightarrow \frac{d y}{d x}=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{\sqrt{1-2 x}}\left\{a+x \tan x+\log \sec x+\frac{1}{x \log x}+\frac{1}{(1-2 x)}\right\}$

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