What is the derivative of $arctan sqrt((1-x)/(1+x))$ ?

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Answer 1 · 2018-05-05T17:49:01

$-1/(2sqrt(1-x^2)), -1 lt x lt 1$ .

Let, $y=arc tansqrt((1-x)/(1+x)$ .

Now, note that, $sqrt((1-x)/(1+x)$ will be meaningful, iff,

$(1-x) ge 0, and (1+x) gt 0, i.e., -1 lt x le 1$ .

Also, the Range of $cos$ function is $[-1,1]$ .

We can, hence, subst., $x=costheta," so that, "theta=arc cos x$ .

Then, we have, $y=arc tansqrt((1-costheta)/(1+costheta))$ ,

$=arc tansqrt((2sin^2(theta/2))/(2cos^2(theta/2))$ ,

$=arc tan(tan(theta/2))$ ,

$=theta/2$ .

$:. y=1/2arc cosx$ ,

$:. dy/dx=1/2*-1/sqrt(1-x^2)=-1/(2sqrt(1-x^2)), -1 lt x lt 1$ .

Enjoy Maths.!

What is the derivative of arctan sqrt((1-x)/(1+x))arctan sqrt((1-x)/(1+x))?