Question

y = tan^-1 under root 1/x - 1 ...then find dy/dx ?????

Answer 1

d (tan^-1√x)/dx=1/1+(√x)^2=1/1+x
is your answer

Answer 2

Hey There!!!

Here we have: 

$y = \tan^{-1} \sqrt{\frac{1}{x}-1}$


Taking tan on both sides: 

$\tan y = \sqrt{\frac{1}{x}-1}$

Now, squaring both sides: 

$\tan^2y = \frac{1}{x}-1$



Now, the overall expression has become fairly easy to differentiate.


Differentiating w.r.t. x : 

$2\tan y \, \sec^2 y \frac{dy}{dx} = -\frac{1}{x^2} + 0 \\ \\ \\ \implies \boxed{\frac{dy}{dx}=-\frac{\cos^2y}{2x^2\tan y} }$
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There is an Alternate Solution as well.


We have: 
$y = \tan^{-1} \sqrt{\frac{1}{x}-1}$

Put $x = \cos^2\theta$

Let us substitute $x=\cos^2\theta$ in 

$y = \tan^{-1} \sqrt{\frac{1}{x}-1} \\ \\ \\ \implies y = \tan^{-1} \sqrt{\frac{1}{\cos^2\theta}-1} \\ \\ \\ \implies y = \tan^{-1} \sqrt{\sec^2\theta - 1} \\ \\ \\ \implies y = \tan^{-1}\sqrt{\tan^2\theta} \\ \\ \\ \implies y = \tan^{-1} \tan\theta \\ \\ \\ \implies y = \theta$ 


Now, we also have: 
$x = \cos^2\theta \\ \\ \implies x = \cos^2 y$ 

Differentiating w.r.t  x 

$1 = 2\cos y (-\sin y) \frac{dy}{dx} \\ \\ \implies \boxed{\frac{dy}{dx}=-\frac{1}{sin 2y}} \\ \\ \\ OR\, \boxed{\frac{dy}{dx}= - cosec \, 2y}$




Hope it helps
Purva
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