Question

Find $\frac{dy}{dx}$

$y=\frac{e^{ax}·\sec x·\log x}{\sqrt{1-2x}}$

Answer 1

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