#### Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 10

$\frac{d y}{d x}=e^{2 \theta}\left[\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{3}+\theta^{2}-1}\right]$

Hint:

Use product rule

Given:

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

Solution:

$x=e^{\theta}\left(\theta+\frac{1}{\theta}\right) \\$

$\frac{d x}{d \theta}=\frac{d}{d \theta}\left[e^{\theta}\left(\theta+\frac{1}{\theta}\right)\right] \\$

\begin{aligned} & &=e^{\theta} \frac{d\left(\theta+\frac{1}{\theta}\right)}{d \theta}+\left(\theta+\frac{1}{\theta}\right) \cdot \frac{d\left(e^{\theta}\right)}{d \theta} \end{aligned}                                                       [Using product rule]

$=e^{\theta}\left[\frac{d \theta}{d \theta}+\frac{d}{d \theta}\left(\frac{1}{\theta}\right)\right]+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta}$                                                                  $\left[\because \frac{d\left(e^{\theta}\right)}{d \theta}=e^{\theta}\right]$

$=e^{\theta}\left[1+\left(-\frac{1}{\theta^{2}}\right)\right]+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta}$                                                                       $\left[\because \frac{d\left(\frac{1}{x}\right)}{d x}=\frac{-1}{x^{2}}\right]$

$=e^{\theta}\left[1-\frac{1}{\theta^{2}}\right]+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta} \\$

$=e^{\theta}\left[\left(1-\frac{1}{\theta^{2}}\right)+\left(\theta+\frac{1}{\theta}\right)\right] \\$

$=e^{\theta}\left[\frac{\theta^{2}-1-\theta^{3}+\theta}{\theta^{2}}\right]$

\begin{aligned} &\\ &\frac{d y}{d \theta}=e^{\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right) \end{aligned}                                                                                           (1)

$y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\$

$\frac{d y}{d \theta}=\frac{d}{d \theta}\left(e^{-\theta} \cdot\left(\theta-\frac{1}{\theta}\right)\right) \\$

\begin{aligned} & &=e^{-\theta}\left(\frac{d \theta}{d \theta}-\frac{d\left(\frac{1}{\theta}\right)}{d \theta}\right]+\left(\theta-\frac{1}{\theta}\right)\left(-e^{-\theta}\right) \end{aligned}

$=e^{-\theta}\left[1-\left(-\frac{1}{\theta^{2}}\right)\right]-e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

$=e^{-\theta}\left[1+\frac{1}{\theta^{2}}\right]-e^{-\theta}\left[\theta-\frac{1}{\theta}\right] \\$

\begin{aligned} & &=e^{-\theta}\left[1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}\right] \end{aligned}

$=e^{-\theta}\left[\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right]$                                                                        $\left[\because L C M\left(\theta, \theta^{2}\right)=\theta^{2}\right]$

$=e^{-\theta}\left[\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right]$

$\frac{d y}{d \theta}=e^{-\theta}\left[\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right]$                                                                                        (2)

Put the values of  $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$  from equation (2) and (1)

$\frac{d y}{d x}=\frac{e^{\theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right)}{e^{-\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right)} \\$

\begin{aligned} & &=e^{\theta} \cdot e^{\theta} \frac{\left(-\theta^{3}+\theta^{2}+\theta+1\right)}{\left(\theta^{3}+\theta^{2}+\theta-1\right)} \end{aligned}

$=e^{2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right) \\$

\begin{aligned} & &\frac{d y}{d x}=e^{2 \theta}\left[\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right] \end{aligned}