Question

Pls answer this question...
( Ans: ____-1____ )
2√(1-x^2)
I need detailed solved answer...​

Answer 1

Answer:

0/0

Step-by-step explanation:

Thats so simple youll just cancel out all the similar variables and turn it into 1

Answer 2

Answer:

$\sf \dfrac{dy}{dx}= \dfrac{-1}{2\sqrt{1-x^2} }$

Step-by-step explanation:

Given:

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

To Find:

dy/dx

Solution:

By given,

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

Let us put x = cos θ

Hence,

$\sf y= \bigg(cot^{-1}\: \dfrac{\sqrt{1+cos\:\theta}-\sqrt{1-cos\:\theta} }{\sqrt{1+cos\:\theta}+\sqrt{1-cos\:\theta} } \bigg)$

Now we know that,

$\sf cos\: \theta=2\:cos^2\: \dfrac{\theta}{2} -1=1-2\:sin^2\:\dfrac{\theta}{2}$

Therefore,

$\sf \implies \bigg(cot^{-1}\: \dfrac{\sqrt{1+2\:cos^2\:\frac{\theta}{2} -1}-\sqrt{1-(1-2\:sin^2\:\frac{\theta}{2} }) }{\sqrt{1+2\:cos^2\:\frac{\theta}{2} -1}+\sqrt{1-(1-2\:sin^2\:\frac{\theta}{2} })} } \bigg)$

$\sf \implies \bigg(cot^{-1}\: \dfrac{\sqrt{2\:cos^2\:\frac{\theta}{2} }-\sqrt{2\:sin^2\:\frac{\theta}{2} } }{\sqrt{2\:cos^2\:\frac{\theta}{2} }+\sqrt{2\:sin^2\:\frac{\theta}{2} }} } \bigg)$

$\sf \implies \bigg(cot^{-1}\: \dfrac{\sqrt{2}\:cos\:\frac{\theta}{2} -\sqrt{2}\:sin\:\frac{\theta}{2} }{\sqrt{2}\:cos\:\frac{\theta}{2} +\sqrt{2}\:sin\:\frac{\theta}{2} } } \bigg)$

Cancelling √2 on both numerator and denominator,

$\sf \implies cot^{-1}\: \bigg(\dfrac{cos\:\frac{\theta}{2} -sin\:\frac{\theta}{2} }{\:cos\:\frac{\theta}{2} +sin\:\frac{\theta}{2} } } \bigg)$

Now taking cos θ/2 common from both numerator and denominator,

$\sf \implies cot^{-1}\: \bigg(\dfrac{cos\:\frac{\theta}{2} (1-\frac{sin\:\frac{\theta}{2} }{cos\:\frac{\theta}{2} } ) }{\:cos\:\frac{\theta}{2}(1 +\frac{sin\:\frac{\theta}{2} }{cos\frac{\theta}{2} })} } \bigg)$

$\sf \implies cot^{-1}\: \bigg(\dfrac{1-\dfrac{sin\:\frac{\theta}{2} }{cos\:\frac{\theta}{2} } }{1 +\dfrac{sin\:\frac{\theta}{2} }{cos\frac{\theta}{2} }} } \bigg)$

$\sf \implies cot^{-1}\bigg(\dfrac{1-tan\:\frac{\theta}{2} }{1+tan\: \frac{\theta}{2} } \bigg)$

$\sf \implies cot^{-1}\bigg(\dfrac{1-tan\: \frac{\pi}{4}\times tan\:\frac{\theta}{2} }{tan\:\frac{\pi}{4} +tan\: \frac{\theta}{2} } \bigg)$

We know that,

$\sf cot(x+y)=\dfrac{1-tan\:x\times tan\:y}{tan\:x+tan\:y}$

Therefore,

$\sf \implies cot^{-1}\bigg(cot(\frac{\pi}{4} +\frac{\theta}{2} )\bigg)$

$\sf \implies \dfrac{\pi}{4} +\dfrac{\theta}{2}$

Now if x = cos θ,

θ = cos⁻¹x

Substitute the value of θ,

$\sf \implies \dfrac{\pi}{4} +\dfrac{cos^{-1}\:x}{2}$

Now differentiating it,

$\sf \dfrac{dy}{dx} =\dfrac{d}{dx} \bigg(\dfrac{\pi}{4} +\dfrac{cos^{-1}\:x}{2} \bigg)$

$\sf \implies \dfrac{-1}{2\sqrt{1-x^2} }$