#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.4 question 18

$\frac{d y}{d x}=\frac{y}{x}$

Hint:

Use quotient rule

Given:

$\sec \left(\frac{x+y}{x-y}\right)=a$

Solution:

\begin{aligned} &\sec \left(\frac{x+y}{x-y}\right)=a \\ &\sec ^{-1} a=\frac{x+y}{x-y} \end{aligned}

Differentiating  $\left[\sec ^{-1} a=\frac{x+y}{x-y}\right]$ w.r.t x

$\frac{d\left(\sec ^{-1} a\right)}{d x}=\frac{d}{d x}\left(\frac{x+y}{x-y}\right)$

$0=\frac{(x-y) \frac{d(x+y)}{d x}-(x+y) \frac{d(x-y)}{d x}}{(x-y)^{2}}$                        [Use quotient rule]

$\frac{(x-y)\left(\frac{d x}{d x}+\frac{d y}{d x}\right)-(x+y)\left(\frac{d x}{d x}-\frac{d y}{d x}\right)}{(x-y)^{2}}=0$

$(x-y)\left(1+\frac{d y}{d x}\right)-(x+y)\left(1-\frac{d y}{d x}\right)=0$

$(x-y)+(x-y) \frac{d y}{d x}-(x+y)+(x+y) \frac{d y}{d x}=0$

$\frac{d y}{d x}[(x-y)+(x+y)]=x+y-(x-y)$

$\frac{d y}{d x}[x-y+x+y]=x+y-x+y$

$\frac{d y}{d x}(2 x)=2 y$

\begin{aligned} &\frac{d y}{d x}=\frac{2 y}{2 x}=\frac{y}{x} \\ &\frac{d y}{d x}=\frac{y}{x} \end{aligned}

Thus, proved