#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 28 maths

$\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$

Hint:

Use chain rule and product rule

Given:

$\cos y=x \cos (a+y)$

Solution:

Differentiate the given equation w.r.t x

$\frac{d \cos y}{d x}=\frac{d(x \cos (a+y))}{d x}$

$\frac{d(\cos y)}{d y} \times \frac{d y}{d x}=x \frac{d(\cos (a+y))}{d x}+\cos (a+y) \cdot \frac{d x}{d x}$                [Use chain rule and product rule]

$(-\sin y) \times \frac{d y}{d x}=x \times\left[\frac{d \cos (a+y)}{d(a+y)} \times \frac{d(a+y)}{d x}\right]+\cos (a+y)$

$-\sin y \frac{d y}{d x}=x \cdot\left(-\sin (a+y) \times\left(\frac{d a}{d x}+\frac{d y}{d x}\right)\right)+\cos (a+y)$

$-\sin y \frac{d y}{d x}=-x \sin (a+y)\left(0+\frac{d y}{d x}\right)+\cos (a+y)$                $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$-\sin y \frac{d y}{d x}=-x \sin (a+y) \frac{d y}{d x}+\cos (a+y)$

$x \sin (a+y) \frac{d y}{d x}-\sin y \frac{d y}{d x}=\cos (a+y)$

$\frac{d y}{d x}(x \sin (a+y)-\sin y)=\cos (a+y)$

Also    $\cos y=x \cos (a+y)$

$x=\frac{\cos y}{\cos (a+y)}$

Put this values of x in the above equation

$\frac{d y}{d x}\left[\frac{\cos y}{\cos (a+y)} \sin (a+y)-\sin y\right]=\cos (a+y)$

$\frac{d y}{d x}\left[\frac{\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)}{\cos (a+y)}\right]=\cos (a+y)$

$\frac{d y}{d x}\left[\frac{\sin ((a+y)-y)}{\cos (a+y)}\right]=\cos (a+y)$                $[\because \sin (A-B)=\sin A \cos B-\cos A \sin B]$

$\frac{d y}{d x} \times\left[\frac{\sin a}{\cos (a+y)}\right]=\cos (a+y)$

\begin{aligned} &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a} \end{aligned}

Thus, proved