Question

y=1+sinx°/1-sinx°differentiate with respect to x

Answer 1

$\huge \mathfrak{solution}$

To Differentiate :

$\bold{y = \frac{1 + \sin x}{1 - \sin x} }$

Here we use the formula :

$\boxed{ \bold{ \star \: \frac{d}{dx} (\frac{u}{v} ) = \frac{v \: \frac{du}{dx} - u \: \frac{dv}{dx} }{ {v}^{2} } }} \\ \\ \star \bold{ \frac{d}{dx} ( \sin x) = \cos x} \\ \\ \star \boxed { \bold{\frac{d}{dx} (constant) = 0}}$

$\implies \: \frac{dy}{dx} = \frac{d}{dx} ( \frac{1 + \sin x}{1 - \sin x} ) \\ \\ \implies \: \frac{dy}{dx} = \frac{(1 - \sin x )\frac{d}{dx} (1 + \sin x) - (1 + \sin x) \frac{d}{dx} (1 - \sin x) }{( {1 - \sin x)}^{2} } \\ \\ \implies \frac{dy}{dx} = \frac{(1 - \sin x)(0 + \cos x) - (1 + \sin x)(0 - \cos x)}{(1 - \sin x) ^{2} } \\ \\ \implies \frac{dy}{dx} = \frac{ \cos x - \cancel{ \sin x \cos x }+ \cos x + \cancel{ \sin x \cos x}}{ {(1 - \sin x)}^{2} } \\ \\ \implies \: \frac{dy}{dx} = \frac{2 \cos x}{( {1 - \sin x)}^{2} }$

Answer 2

solution

To Differentiate :

\begin{lgathered}\bold{y = \frac{1 + \sin x}{1 - \sin x} }\\\end{lgathered}

y=

1−sinx

1+sinx

Here we use the formula :

\begin{lgathered}\boxed{ \bold{ \star \: \frac{d}{dx} (\frac{u}{v} ) = \frac{v \: \frac{du}{dx} - u \: \frac{dv}{dx} }{ {v}^{2} } }} \\ \\ \star \bold{ \frac{d}{dx} ( \sin x) = \cos x} \\ \\ \star \boxed { \bold{\frac{d}{dx} (constant) = 0}}\end{lgathered}

⋆

dx

d

(

v

u

)=

v

2

v

dx

du

−u

dx

dv

⋆

dx

d

(sinx)=cosx

⋆

dx

d

(constant)=0

\begin{lgathered}\implies \: \frac{dy}{dx} = \frac{d}{dx} ( \frac{1 + \sin x}{1 - \sin x} ) \\ \\ \implies \: \frac{dy}{dx} = \frac{(1 - \sin x )\frac{d}{dx} (1 + \sin x) - (1 + \sin x) \frac{d}{dx} (1 - \sin x) }{( {1 - \sin x)}^{2} } \\ \\ \implies \frac{dy}{dx} = \frac{(1 - \sin x)(0 + \cos x) - (1 + \sin x)(0 - \cos x)}{(1 - \sin x) ^{2} } \\ \\ \implies \frac{dy}{dx} = \frac{ \cos x - \cancel{ \sin x \cos x }+ \cos x + \cancel{ \sin x \cos x}}{ {(1 - \sin x)}^{2} } \\ \\ \implies \: \frac{dy}{dx} = \frac{2 \cos x}{( {1 - \sin x)}^{2} }\end{lgathered}

⟹

dx

dy

=

dx

d

(

1−sinx

1+sinx

)

⟹

dx

dy

=

(1−sinx)

2

(1−sinx)

dx

d

(1+sinx)−(1+sinx)

dx

d

(1−sinx)

⟹

dx

dy

=

(1−sinx)

2

(1−sinx)(0+cosx)−(1+sinx)(0−cosx)

⟹

dx

dy

=

(1−sinx)

2

cosx−

sinxcosx

+cosx+

sinxcosx

⟹

dx

dy

=

(1−sinx)

2

2cosx