#### Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 15

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

Hint:

Use chain rule and properties of logarithm

Given:

\begin{aligned} &x=e^{\cos 2 t} \\ &y=e^{\sin 2 t} \end{aligned}

Solution:

$x=e^{\cos 2 t} \\$

$\frac{d x}{d t}=\frac{d\left(e^{\cos 2 t}\right)}{d t} \\$

\begin{aligned} &=\frac{d\left(e^{\cos 2 t}\right)}{d(\cos 2 t)} \times \frac{d(\cos 2 t)}{d(2 t)} \times \frac{d(2 t)}{d t} \end{aligned}                                                                                           [Using chain rule]

$=e^{\cos 2 t} \times(-\sin 2 t) \times 2 \\$                                                                  $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x} \frac{d(\cos \theta)}{d \theta}=-\sin \theta\right]$

\begin{aligned} & &\frac{d x}{d t}=-2 \sin 2 t e^{\cos 2 t} \end{aligned}                                                                                                                                                        (1)

$y=e^{\sin 2 t} \\$

$\frac{d y}{d t}=\frac{d\left(e^{\sin 2 t}\right)}{d t} \\$

\begin{aligned} & &=\frac{d\left(e^{\sin 2 t}\right)}{d(\sin 2 t)} \times \frac{d(\sin 2 t)}{d(2 t)} \times \frac{d(2 t)}{d t} \end{aligned}                                                            [Using chain rule]

$=e^{\sin 2 t} \times \cos 2 t \times 2 \\$                                                                                         $\left[\because \frac{d e^{x}}{d x}=e^{x}, \frac{d \sin \theta}{d \theta}=\cos \theta\right]$

$\frac{d y}{d t}=2 \cos 2 t e^{\sin 2 t} \\$                                                                                                                                                          (2)

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

Putting the value of  $\frac{d x}{d t} \text { and } \frac{d y}{d t}$  from the equation (1) and (2) respectively

$=\frac{2 \cos 2 t \cdot e^{\sin 2 t}}{-2 \sin 2 t \cdot e^{\sin 2 t}}$

\begin{aligned} &\frac{d y}{d x}=-\frac{\cos 2 t \cdot e^{\sin 2 t}}{\sin 2 t \cdot e^{\cos 2 t}} \\ & \end{aligned}                                                                                                                                                      (3)

$x=e^{\cos 2 t}$

Take log on both sides

$\log x=\log \left(e^{\cos 2 t}\right) \\$

$\log x=\cos 2 t \cdot \log e \\$                                                                                                      $\left[\because \log a^{m}=m \log a\right]$

\begin{aligned} & &\log x=\cos 2 t \end{aligned}                                                                                                  $[\because \log e=1]$

Also

$y=e^{\sin 2 t}$

Take log on both side

\begin{aligned} &\log y=\log \left(e^{\sin 2 t}\right) \\ &\log y=\sin 2 t \cdot \log e \\ &\log y=\sin 2 t \end{aligned}

Put  $e^{\cos 2 t}=x, \cos 2 t=\log x, e^{\sin 2 t}=y, \sin 2 t=\log y$  in equation (3)

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

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

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