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

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

Answers (1)

Answer:

            \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}

 

Posted by

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