Find dy dx in the following cases: 4x+3y= (4x-3y ) - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$

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Answer

We have, $4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(4\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(3\text{y}\big)=\frac{\text{d}}{\text{dx}}\big\{\log\big(4\text{x}-3\text{y}\big)\big\}$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\frac{\text{d}}{\text{dx}}\big(4\text{x}-3\text{y}\big)$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\Big(4-3\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}+\frac{3}{\big(4\text{x}-3\text{y}\big)}\frac{\text{dy}}{\text{dx}}=\frac{4}{\big(4\text{x}-3\text{y}\big)}-4$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{1+\frac{1}{(4\text{x}-3\text{y})}\Big\}=4\Big\{\frac{1}{(4\text{x}-3\text{y})}-1\Big\}$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{\frac{4\text{x}-3\text{y}+1}{(4\text{x}+3\text{y})}\Big\}=4\Big\{\frac{1-4\text{x}-3\text{y}}{4\text{x}-3\text{y}}\Big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big\{\frac{1-4\text{x}+3\text{y}}{(4\text{x}-3\text{y})}\Big\}\Big(\frac{4\text{x}-3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big(\frac{1-4\text{x}+3\text{y}}{4\text{x}-3\text{y}+1}\Big)$

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