Question

Differentiate w.r.t.x
​

Answer 1

Answer:

$\large\boxed{\sf{\dfrac{2 \cos(x) }{ {(1 - \sin x) }^{2} } }}$

Step-by-step explanation:

To differentiate ,

$\dfrac{1 + \sin(x) }{1 - \sin(x) }$

We know that,

• $\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx}g(x) }{ {(g(x))}^{2} }$

Therefore, we will get,

$= \dfrac{(1 - \sin x) \frac{d}{dx} (1 + \sin x) - (1 + \sin x) \frac{d}{dx}(1 - \sin x) }{ {(1 - \sin x) }^{2} }$

But, we know that,

• $\dfrac{d}{dx} \sin(x) = \cos(x)$

• $\dfrac{d}{dx} c = 0$

Where, c is any constant.

Therefore, we will get,

$= \dfrac{( 1- \sin x) \cos(x) - ( 1 + \sin x)( - \cos x) }{ {(1 - \sin x) }^{2} } \\ \\ = \frac{ \cos(x) - \sin(x) \cos(x) + \cos(x) + \sin(x) \cos(x) }{ {(1 - \sin x) }^{2} } \\ \\ = \dfrac{2 \cos(x) }{ {(1 - \sin x) }^{2} }$

Hence, the required value is $\bold{\dfrac{2 \cos(x) }{ {(1 - \sin x) }^{2} } }$

Answer 2

ANSWER

=> tan(π/4+x/2) sec^2 (π/4+x/2)

FORMULAS OF TRIGNOMETRY

AND LOGIC FOR SOLVING QUESTION

1 + sinx = sin^(x/2) +cos ^2(x/2) + 2sin(x/2) cos(x/2)

= (cosx/2 + sinx)

logic is first reduce function to simple expression and then differentiate

SOLUTION

USE FORMULA

WE GET

(cosx/2 + sinx/2) ^2

( cosx/2 - sinx/2)^2

divide numerator and denominator by cos(x/2)

inside the braces of square

we get

(1+tanx/2) ^2

(1-tanx/2) ^2

we write 1 = tan π/4

and we know tan(x+y) = tanx + tany / 1- tanx tany

so it reduces to

[tan(π/4+x/2) ]^2

now differentiate it wrt x

we get

=>2 tan(π/4+x/2) d(tan(π/4+x/2))

dx

=>2 tan(π/4+x/2) sec^2 (π/4+x/2) (d(x/2)

dx

=>2 tan(π/4+x/2) sec^2 (π/4+x/2) ×1/2

=> tan(π/4+x/2) sec^2 (π/4+x/2)

I use different method because in competitive exams options are set as different to confuse student

both answeres are convettable

hope it helps