Find dy dx y= e^ ax 1-2x - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}$

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Answer

Here,
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}\ .....(\text{i})$
$\Rightarrow\text{y}=\frac{\text{e}^{\text{ax}}\times\sec^\text{x}\times\log\text{x}}{(1-2\text{x})^\frac{1}{2}}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{\text{ax}}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \begin{bmatrix} \text{Since}, \log\Big(\frac{\text{A}}{\text{B}}\Big)=\log\text{A}-\log\text{B},\\ \log(\text{AB})=\log\text{A}+\log\text{B} \end{bmatrix}$
$\log\text{y}=\text{ax}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a and }\log_\text{e}\text{e}=1\big]$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{ax})+\frac{\text{d}}{\text{dx}}(\log\sec\text{x})+\frac{\text{d}}{\text{dx}}(\log\log\text{x})-\frac{1}{2}\log(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{1}{\sec\text{x}}\frac{\text{d}}{\text{dx}}(\sec\text{x})+\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)\frac{\text{d}}{\text{dx}}(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{\sec\text{x}\tan\text{x}}{\sec\text{x}}+\frac{1}{(\log\text{x})}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)(-2)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{ax}}\sec\text{x}\log\text{x}}{\sqrt{1-2\text{x}}}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
[Using equation (i)]

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