If \[y = \log \sqrt{\tan x}, \text{ write } \frac{dy}{dx} \] ?

\[\text{ We have, y } = \log\sqrt{\tan x}\] \[ \Rightarrow y = \log \left( \tan x \right)^\frac{1}{2} \] \[ \Rightarrow y = \frac{1}{2}\log \tan x \left[ \because \log a^b = b\log a \right]\] \[\Rightarrow \frac{dy}{dx} = \frac{1}{2} \times \frac{1}{\tan x}\frac{d}{dx}\left( \tan x \right)\] \[ \Rightarrow \frac{dy}{dx} = \frac{1}{2} \times \frac{1}{\tan x}\left( \sec^2 x \right)\] \[ \Rightarrow \frac{dy}{dx} = \frac{1}{2\frac{\sin x}{\cos x} \times \cos^2 x}\] \[ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sin x \cos x}\] \[ \Rightarrow \frac{dy}{dx} = \frac{1}{\sin2x}\] \[ \Rightarrow \frac{dy}{dx} = cosec \ 2x\]

If Y = Log √ Tan X , Write D Y D X ? - Mathematics

Question

If $y = \log \sqrt{\tan x}, write \frac{d y}{d x}$ ?

Solution

$We have, y = \log \sqrt{\tan x}$

$\Rightarrow y = \log {(\tan x)}^{\frac{1}{2}}$

$\Rightarrow y = \frac{1}{2} \log \tan x [∵ \log a^{b} = b \log a]$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2} \times \frac{1}{\tan x} \frac{d}{d x} (\tan x)$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2} \times \frac{1}{\tan x} (\sec^{2} x)$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 \frac{\sin x}{\cos x} \times \cos^{2} x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 \sin x \cos x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\sin 2 x}$

$\Rightarrow \frac{d y}{d x} = c o s e c 2 x$

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Simple Problems on Applications of Derivatives

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Q 19 Q 18 Q 20

Chapter 11: Differentiation - Exercise 11.09 [Page 118]

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Chapter 11 Differentiation
Exercise 11.09 | Q 19 | Page 118

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If Y = Log √ Tan X , Write D Y D X ? - Mathematics

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