Given Y=sin(2sin^-1(x)), find dy/dx ?

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Answer 1 · 2018-06-26T16:21:30

$dy/dx = (2(1-2x^2))/sqrt(1-x^2)$

Let $t=arcsinx$ , which is defined for $x in (-pi/2,pi/2)$

Then:

$sint = x$

and

$cost = sqrt(1-x^2)$

because for $x in (-pi/2,pi/2)$ the cosine is positive.

So:

$y= sin(2arcsinx) = sin2t = 2sintcost = 2xsqrt(1-x^2)$

and using the product rule:

$dy/dx = 2xd/dx(sqrt(1-x^2)) +2sqrt(1-x^2)$

$dy/dx = 2x(-(2x)/(2sqrt(1-x^2))) +2sqrt(1-x^2)$

$dy/dx = -(2x^2)/sqrt(1-x^2) +2sqrt(1-x^2)$

$dy/dx = (-2x^2+2-2x^2)/sqrt(1-x^2)$

$dy/dx = (2(1-2x^2))/sqrt(1-x^2)$

Answer 2 · 2018-06-26T20:26:28

$y=sin(2sin^-1(x))$

$underbrace(sin^(-1) y)_(alpha)= 2 underbrace(sin^-1(x))_(beta) qquad qquad bb(alpha = 2 beta)$

$implies y = 2 x sqrt(1-x^2)$

Apply the product rule etc to find:

$y'= d/dx(2 x sqrt(1 - x^2)) = 2*(1 - 2 x^2)/sqrt(1 - x^2)$