Question

differentiate y = log √1+sinx/1 - sinx​

Answer 1

Answer:

$\sf \dfrac{dy}{dx} =sec\:x$

Step-by-step explanation:

Given:

$\sf y=log\:\sqrt{\dfrac{1+sin\:x}{1-sin\:x} }$

To Find:

dy/dx

Solution:

$\sf y=log\:\sqrt{\dfrac{1+sin\:x}{1-sin\:x} }$

$\sf \implies log\: \bigg(\dfrac{1+sin\:x}{1-sin\:x} \bigg)^{\dfrac{1}{2}}$

We know that,

$\sf log\: (a^b)=b\:log\:a$

Hence,

$\sf \implies \dfrac{1}{2} \:log\: \bigg(\dfrac{1+sin\:x}{1-sin\:x} \bigg)$

Also,

$\sf log\: \bigg(\dfrac{a}{b} \bigg)=log\:a-log\:b$

$\implies\sf \dfrac{1}{2} \bigg(log\:(1+sin\:x)-log(1-sin\:x)\bigg)$

$\sf \implies \dfrac{log(1+sin\:x)}{2} -\dfrac{log(1-sin\:x)}{2}$

Now differentiate with respect to x,

$\sf \dfrac{dy}{dx} =\dfrac{d}{dx} \bigg(\dfrac{log(1+sin\:x)}{2} \bigg)-\dfrac{d}{dx}\bigg(\dfrac{log(1-sin\:x)}{2} \bigg)$

$\implies \sf \dfrac{1}{2} \: \dfrac{d}{dx}\: log(1+sin\:x)-\dfrac{1}{2}\:\dfrac{d}{dx}\:log\:(1-sin\:x)$

Now by using chain rule,

$\implies \sf \dfrac{1}{2}\times \dfrac{1}{1+sin\:x} \:\dfrac{d}{dx} (1+sin\:x)-\dfrac{1}{2}\times \dfrac{1}{1-sin\:x} \: \dfrac{d}{dx} (1-sin\:x)$

$\implies \sf \dfrac{1}{2(1+sin\:x)}\times cos\:x-\dfrac{1}{2(1-sin\:x)} \times -cos\:x$

$\implies \sf \dfrac{1}{2(1+sin\:x)}\times cos\:x+\dfrac{1}{2(1-sin\:x)} \times cos\:x$

Taking 1/2 × cos x common,

$\sf \implies \dfrac{1}{2} \times cos\:x\bigg(\dfrac{1}{1+sin\:x} +\dfrac{1}{1-sin\:x} \bigg)$

$\implies \sf \dfrac{1}{2} \times cos\:x\bigg(\dfrac{1-sin\:x+1+sin\:x}{1-sin^2\:x} \bigg)$

We know that 1 - sin² x = cos² x

Hence,

$\implies \sf \dfrac{1}{2} \times cos\:x\bigg(\dfrac{2}{cos^2\:x} \bigg)$

$\sf \implies \dfrac{1}{cos\:x}=sec\:x$

Answer 2

Answer:

$Answer:</p><p></p><p>\sf \dfrac{dy}{dx} =sec\:x </p><p>dx</p><p>dy</p><p> </p><p> =secx</p><p></p><p>Step-by-step explanation:</p><p></p><p>Given:</p><p></p><p>\sf y=log\:\sqrt{\dfrac{1+sin\:x}{1-sin\:x} }y=log </p><p>1−sinx</p><p>1+sinx</p><p> </p><p> </p><p> </p><p> </p><p></p><p></p><p>To Find:</p><p></p><p>dy/dx</p><p></p><p>Solution:</p><p></p><p>\sf y=log\:\sqrt{\dfrac{1+sin\:x}{1-sin\:x} }y=log </p><p>1−sinx</p><p>1+sinx</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\sf \implies log\: \bigg(\dfrac{1+sin\:x}{1-sin\:x} \bigg)^{\dfrac{1}{2}}⟹log( </p><p>1−sinx</p><p>1+sinx</p><p> </p><p> ) </p><p>2</p><p>1</p><p> </p><p> </p><p> </p><p></p><p>We know that,</p><p></p><p>\sf log\: (a^b)=b\:log\:alog(a </p><p>b</p><p> )=bloga</p><p></p><p>Hence,</p><p></p><p>\sf \implies \dfrac{1}{2} \:log\: \bigg(\dfrac{1+sin\:x}{1-sin\:x} \bigg)⟹ </p><p>2</p><p>1</p><p> </p><p> log( </p><p>1−sinx</p><p>1+sinx</p><p> </p><p> )</p><p></p><p>Also,</p><p></p><p>\sf log\: \bigg(\dfrac{a}{b} \bigg)=log\:a-log\:blog( </p><p>b</p><p>a</p><p> </p><p> )=loga−logb</p><p></p><p>\implies\sf \dfrac{1}{2} \bigg(log\:(1+sin\:x)-log(1-sin\:x)\bigg)⟹ </p><p>2</p><p>1</p><p> </p><p> (log(1+sinx)−log(1−sinx))</p><p></p><p>\sf \implies \dfrac{log(1+sin\:x)}{2} -\dfrac{log(1-sin\:x)}{2}⟹ </p><p>2</p><p>log(1+sinx)</p><p> </p><p> − </p><p>2</p><p>log(1−sinx)</p><p> </p><p> </p><p></p><p>Now differentiate with respect to x,</p><p></p><p>\sf \dfrac{dy}{dx} =\dfrac{d}{dx} \bigg(\dfrac{log(1+sin\:x)}{2} \bigg)-\dfrac{d}{dx}\bigg(\dfrac{log(1-sin\:x)}{2} \bigg) </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>dx</p><p>d</p><p> </p><p> ( </p><p>2</p><p>log(1+sinx)</p><p> </p><p> )− </p><p>dx</p><p>d</p><p> </p><p> ( </p><p>2</p><p>log(1−sinx)</p><p> </p><p> )</p><p></p><p>\implies \sf \dfrac{1}{2} \: \dfrac{d}{dx}\: log(1+sin\:x)-\dfrac{1}{2}\:\dfrac{d}{dx}\:log\:(1-sin\:x)⟹ </p><p>2</p><p>1</p><p> </p><p> </p><p>dx</p><p>d</p><p> </p><p> log(1+sinx)− </p><p>2</p><p>1</p><p> </p><p> </p><p>dx</p><p>d</p><p> </p><p> log(1−sinx)</p><p></p><p>Now by using chain rule,</p><p></p><p>\implies \sf \dfrac{1}{2}\times \dfrac{1}{1+sin\:x} \:\dfrac{d}{dx} (1+sin\:x)-\dfrac{1}{2}\times \dfrac{1}{1-sin\:x} \: \dfrac{d}{dx} (1-sin\:x)⟹ </p><p>2</p><p>1</p><p> </p><p> × </p><p>1+sinx</p><p>1</p><p> </p><p> </p><p>dx</p><p>d</p><p> </p><p> (1+sinx)− </p><p>2</p><p>1</p><p> </p><p> × </p><p>1−sinx</p><p>1</p><p> </p><p> </p><p>dx</p><p>d</p><p> </p><p> (1−sinx)</p><p></p><p>\implies \sf \dfrac{1}{2(1+sin\:x)}\times cos\:x-\dfrac{1}{2(1-sin\:x)} \times -cos\:x⟹ </p><p>2(1+sinx)</p><p>1</p><p> </p><p> ×cosx− </p><p>2(1−sinx)</p><p>1</p><p> </p><p> ×−cosx</p><p></p><p>\implies \sf \dfrac{1}{2(1+sin\:x)}\times cos\:x+\dfrac{1}{2(1-sin\:x)} \times cos\:x⟹ </p><p>2(1+sinx)</p><p>1</p><p> </p><p> ×cosx+ </p><p>2(1−sinx)</p><p>1</p><p> </p><p> ×cosx</p><p></p><p>Taking 1/2 × cos x common,</p><p></p><p>\sf \implies \dfrac{1}{2} \times cos\:x\bigg(\dfrac{1}{1+sin\:x} +\dfrac{1}{1-sin\:x} \bigg)⟹ </p><p>2</p><p>1</p><p> </p><p> ×cosx( </p><p>1+sinx</p><p>1</p><p> </p><p> + </p><p>1−sinx</p><p>1</p><p> </p><p> )</p><p></p><p>\implies \sf \dfrac{1}{2} \times cos\:x\bigg(\dfrac{1-sin\:x+1+sin\:x}{1-sin^2\:x} \bigg)⟹ </p><p>2</p><p>1</p><p> </p><p> ×cosx( </p><p>1−sin </p><p>2</p><p> x</p><p>1−sinx+1+sinx</p><p> </p><p> )</p><p></p><p>We know that 1 - sin² x = cos² x</p><p></p><p>Hence,</p><p></p><p>\implies \sf \dfrac{1}{2} \times cos\:x\bigg(\dfrac{2}{cos^2\:x} \bigg)⟹ </p><p>2</p><p>1</p><p> </p><p> ×cosx( </p><p>cos </p><p>2</p><p> x</p><p>2</p><p> </p><p> )</p><p></p><p>\sf \implies \dfrac{1}{cos\:x}=sec\:x⟹ </p><p>cosx</p><p>1</p><p> </p><p> =secx$