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Name: Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 15
Uploaded: Sun, 2021-08-15 15:14
Description: Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 15

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

Answers (1)

Answer:

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

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]