Question

differentiate w.r.t.x​

Answer 1

AnswEr :

Given Expression,

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

We know that,

log(a/b) = log(a) - log(b)

Thus,

$\implies \sf \: y = log( \sqrt{1 - sin \: x} ) - log( \sqrt{1 + sin \: x} )$

We know that,

$\sf \star \: \boxed{ \boxed{ \sf \dfrac{d( log(x)) }{dx} = \dfrac{1}{x} }}$

Differentiating the above expression w.r.t x,

$\longrightarrow \: \sf \: \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 - sin \: x} } \times \bigg(\dfrac{d( \sqrt{1 - sin \: x)} }{dx} \bigg) - \dfrac{1}{ \sqrt{1 + sin \: x} } \times \bigg( \dfrac{d( \sqrt{1 + sin \: x}) }{dx} \bigg)$

We know that,

$\sf \star \: \boxed{ \boxed{ \sf \dfrac{d( \sqrt{x} ) }{dx} = \dfrac{1}{2 \sqrt{x} } }}$

Using Chain Rule,

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

We know that,

$\sf \star \: \boxed{ \boxed{ \sf \dfrac{d( sin \: x) }{dx} = cos \: x }}$

Thus,

$\longrightarrow \: \sf \: \dfrac{dy}{dx} = \dfrac{ - cos \: x}{2( \sqrt{1 - sin \: x}) {}^{2} } + \dfrac{cos \: x}{2( \sqrt{1 + sin \: x} ) {}^{2} } \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = \dfrac{cos \: x}{2} \bigg[ \frac{1}{1 + sin \: x} - \dfrac{1}{1 - sin \: x} \bigg] \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = \dfrac{cos \: x}{2} \bigg[ \frac{(1 - sin \: x) - (1 + sin \: x)}{1 - {sin}^{2}x } \bigg] \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = \dfrac{cos \: x}{ \cancel{2}} \times \bigg( - \dfrac{ \cancel{2}sin \: x}{cos {}^{2} x } \bigg) \\ \\ \longrightarrow \: \sf \: \dfrac{dy}{dx} = - \dfrac{sin \: x}{cos \: x } \times \cancel{ \dfrac{cos \: x}{cos \: x} }\\ \\ \longrightarrow \boxed{ \boxed{ \sf \: \dfrac{dy}{dx} = - tan \: x}}$

Derivative of the above expression is - tan(x)