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Question
Find \[\frac{dy}{dx}\] in the following case \[4x + 3y = \log \left( 4x - 3y \right)\] ?
Solution
We have,
\[4x + 3y = \log\left( 4x - 3y \right)\]
Differentiating with respect to x, we get,
\[\frac{d}{dx}\left( 4x \right) + \frac{d}{dx}\left( 3y \right) = \frac{d}{dx}\left\{ \log\left( 4x - 3y \right) \right\}\]
\[ \Rightarrow 4 + 3\frac{d y}{d x} = \frac{1}{\left( 4x - 3y \right)}\frac{d}{dx}\left( 4x - 3y \right) \]
\[ \Rightarrow 4 + 3\frac{d y}{d x} = \frac{1}{\left( 4x - 3y \right)}\left( 4 - 3\frac{d y}{d x} \right)\]
\[ \Rightarrow 3\frac{d y}{d x} + \frac{3}{\left( 4x - 3y \right)}\frac{d y}{d x} = \frac{4}{\left( 4x - 3y \right)} - 4\]
\[ \Rightarrow 3\frac{d y}{d x}\left\{ 1 + \frac{1}{\left( 4x - 3y \right)} \right\} = 4\left\{ \frac{1}{\left( 4x - 3y \right)} - 1 \right\}\]
\[ \Rightarrow 3\frac{d y}{d x}\left\{ \frac{4x - 3y + 1}{\left( 4x - 3y \right)} \right\} = 4\left\{ \frac{1 - 4x + 3y}{\left( 4x - 3y \right)} \right\}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{4}{3}\left\{ \frac{1 - 4x + 3y}{\left( 4x - 3y \right)} \right\}\left( \frac{4x - 3y}{4x - 3y + 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{4}{3}\left( \frac{1 - 4x + 3y}{4x - 3y + 1} \right)\]
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