Question

Find $\frac{dy}{dx}$ in each of the following cases:

$4x+3y=\log \left(4x-3y\right)$

Answer 1

$\mathrm{We}\mathrm{have},4x+3y=\log \left(4x-3y\right)$
Differentiating with respect to x, we get,

$$\begin{gathered} \frac{d}{dx}\left(4x\right)+\frac{d}{dx}\left(3y\right)=\frac{d}{dx}\left\{\log \left(4x-3y\right)\right\} \\ \Rightarrow 4+3\frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{\left(4x-3y\right)}\frac{d}{dx}\left(4x-3y\right) \\ \Rightarrow 4+3\frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{\left(4x-3y\right)}\left(4-3\frac{dy}{dx}\right) \\ \Rightarrow 3\frac{\mathit{d}y}{\mathit{d}x}+\frac{3}{\left(4x-3y\right)}\frac{\mathit{d}y}{\mathit{d}x}=\frac{4}{\left(4x-3y\right)}-4 \\ \Rightarrow 3\frac{\mathit{d}y}{\mathit{d}x}\left\{1+\frac{1}{\left(4x-3y\right)}\right\}=4\left\{\frac{1}{\left(4x-3y\right)}-1\right\} \\ \Rightarrow 3\frac{\mathit{d}y}{\mathit{d}x}\left\{\frac{4x-3y+1}{\left(4x-3y\right)}\right\}=4\left\{\frac{1-4x+3y}{\left(4x-3y\right)}\right\} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{4}{3}\left\{\frac{1-4x+3y}{\left(4x-3y\right)}\right\}\left(\frac{4x-3y}{4x-3y+1}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{4}{3}\left(\frac{1-4x+3y}{4x-3y+1}\right) \end{gathered}$$