#### Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 4

$\frac{dy}{dx}=\cot \theta$

Hint:

Use product rule

Given:

$x=a e^{\theta}(\sin \theta-\cos \theta)$

\begin{aligned}&y=a e^{\theta}(\sin \theta+\cos \theta) \end{aligned}

Solution:

$x=a e^{\theta}(\sin \theta-\cos \theta)$

\begin{aligned} & \\ &\frac{d x}{d \theta}=a e^{\theta} \frac{d(\sin \theta-\cos \theta)}{d \theta}+(\sin \theta-\cos \theta) \frac{d\left(a e^{\theta}\right)}{d \theta} \end{aligned}                                        [Use product rule]

\begin{aligned} &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) a \cdot \frac{d e^{\theta}}{d \theta} \\ &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) \cdot a \cdot e^{\theta} \\ &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta+\sin \theta-\cos \theta) \end{aligned}

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

$\frac{d y}{d \theta}=a e^{\theta} \cdot \frac{d(\sin \theta+\cos \theta)}{d \theta}+(\sin \theta+\cos \theta) \cdot \frac{d\left(a e^{\theta}\right)}{d \theta}$                         [Using product rule]

\begin{aligned} &=a e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) a \cdot \frac{d e^{\theta}}{d \theta} \\ &=a e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) a e^{\theta} \\ &=a e^{\theta}(\cos \theta-\sin \theta+\sin \theta+\cos \theta) \\ &\frac{d y}{d \theta}=2 a e^{\theta} \cos \theta \end{aligned}

So,

$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2 a e^{\theta} \cos \theta}{2 a e^{\theta} \sin \theta}=\frac{\cos \theta}{\sin \theta}=\cot \theta$