#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 4 maths

$\frac{4}{3}\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$

Hint:

Use chain rule to find the differentiation

Given:

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

Solution:

Differentiate the given equation w.r.t x

$\frac{d}{d x}(4 x+3 y)=\frac{d}{d x}(\log (4 x-3 y))$

$\frac{d}{d x}(4 x)+\frac{d}{d x}(3 y)=\frac{d(\log (4 x-3 y))}{d(4 x-3 y)} \times \frac{d(4 x-3 y)}{d x}$

[Using the chain  $\frac{d(f(x+y))}{d x}=\frac{d(f(x+y))}{d(x+y)} \times \frac{d(x+y)}{d x}$    ]

$4 \frac{d x}{d x}+3 \frac{d y}{d x}=\frac{1}{(4 x-3 y)} \times\left[\frac{d(4 x)}{d x}-\frac{d(3 y)}{d x}\right] \quad\left[\because \frac{d(\log x)}{d x}=\frac{1}{x}\right]$

$4+3 \frac{d y}{d x}=\frac{1}{4 x-3 y} \cdot\left[4-\frac{d(3 y)}{d y} \times \frac{d y}{d x}\right]$

$4+3 \frac{d y}{d x}=\frac{1}{4 x-3 y} \cdot\left[4-3 \times \frac{d y}{d x}\right]$

$4+3 \frac{d y}{d x}=\frac{4}{4 x-3 y}-\frac{3}{4 x-3 y} \frac{d y}{d x}$

$3 \frac{d y}{d x}+\frac{3}{4 x-3 y} \cdot \frac{d y}{d x}=\frac{4}{4 x-3 y}-4$

$3 \frac{d y}{d x}\left(1+\frac{1}{4 x-3 y}\right)=\frac{4-4(4 x-3 y)}{4 x-3 y}$

$3 \frac{d y}{d x} \times\left(\frac{4 x-3 y+1}{4 x-3 y}\right)=\frac{4-16 x+12 y}{4 x-3 y}$

$\frac{d y}{d x}=\frac{4-16 x+12 y}{(4 x-3 y)} \times \frac{4 x-3 y}{(4 x-3 y+1)} \times \frac{1}{3}$

$\frac{d y}{d x}=\frac{1}{3} \cdot \frac{(4-16 x+12 y)}{(4 x-3 y+1)}$

$\frac{d y}{d x}=\frac{4}{3} \cdot\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$

Hence

$\frac{d y}{d x}=\frac{4}{3}\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$  Is the required answer