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प्रश्न
Find \[\frac{dy}{dx}\] \[y = \frac{e^{ax} \cdot \sec x \cdot \log x}{\sqrt{1 - 2x}}\] ?
उत्तर
\[\text{ We have, y } = \frac{e^{ax} \sec x \log x}{\sqrt{1 - 2x}} . . . \left( i \right)\]
\[ \Rightarrow y = \frac{e^{ax} \sec x \log x}{\left( 1 - 2x \right)^\frac{1}{2}}\]
Taking log on both sides
\[\log y = \log e^{ax} + logsec x + \log \log x - \frac{1}{2}\log\left( 1 - 2x \right) \]
\[ \Rightarrow \log y = ax + \log\left( \sec x \right) + \log\left( \log x \right) - \frac{1}{2}\log\left( 1 - 2x \right) \]
Differentiating with respect to x using chain rule,
\[\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left( ax \right) + \frac{d}{dx}\left( \log \sec x \right) + \frac{d}{dx}\left( \log \log x \right) - \frac{1}{2}\log\left( 1 - 2x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = a + \frac{1}{\sec x}\frac{d}{dx}\left( \sec x \right) + \frac{1}{\log x}\frac{d}{dx}\left( \log x \right) - \frac{1}{2}\left( \frac{1}{1 - 2x} \right)\frac{d}{dx}\left( 1 - 2x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = a + \frac{\sec x \tan x}{\sec x} + \frac{1}{\left( \log x \right)}\left( \frac{1}{x} \right) - \frac{1}{2}\left( \frac{1}{1 - 2x} \right)\left( - 2 \right)\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ a + \tan x + \frac{1}{x \log x} + \frac{1}{1 - 2x} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^{ax} \sec x \log x}{\sqrt{1 - 2x}}\left[ a + \tan x + \frac{1}{x \log x} + \frac{1}{1 - 2x} \right] \left[ \text{ Using equation }\left( i \right) \right]\]
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