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

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Answer 1 · 2018-06-26T16:21:30

$\frac{d y}{d x} = \frac{2 (1 - 2 x^{2})}{\sqrt{1 - x^{2}}}$ $dy/dx = (2(1-2x^2))/sqrt(1-x^2)$

Let $t = arcsin x$ $t=arcsinx$ , which is defined for $x \in (- \frac{π}{2}, \frac{π}{2})$ $x in (-pi/2,pi/2)$

Then:

$sin t = x$ $sint = x$

and

$cos t = \sqrt{1 - x^{2}}$ $cost = sqrt(1-x^2)$

because for $x \in (- \frac{π}{2}, \frac{π}{2})$ $x in (-pi/2,pi/2)$ the cosine is positive.

So:

$y = sin (2 arcsin x) = sin 2 t = 2 sin t cos t = 2 x \sqrt{1 - x^{2}}$ $y= sin(2arcsinx) = sin2t = 2sintcost = 2xsqrt(1-x^2)$

and using the product rule:

$\frac{d y}{d x} = 2 x \frac{d}{d x} (\sqrt{1 - x^{2}}) + 2 \sqrt{1 - x^{2}}$ $dy/dx = 2xd/dx(sqrt(1-x^2)) +2sqrt(1-x^2)$

$\frac{d y}{d x} = 2 x (- \frac{2 x}{2 \sqrt{1 - x^{2}}}) + 2 \sqrt{1 - x^{2}}$ $dy/dx = 2x(-(2x)/(2sqrt(1-x^2))) +2sqrt(1-x^2)$

$\frac{d y}{d x} = - \frac{2 x^{2}}{\sqrt{1 - x^{2}}} + 2 \sqrt{1 - x^{2}}$ $dy/dx = -(2x^2)/sqrt(1-x^2) +2sqrt(1-x^2)$

$\frac{d y}{d x} = \frac{- 2 x^{2} + 2 - 2 x^{2}}{\sqrt{1 - x^{2}}}$ $dy/dx = (-2x^2+2-2x^2)/sqrt(1-x^2)$

$\frac{d y}{d x} = \frac{2 (1 - 2 x^{2})}{\sqrt{1 - x^{2}}}$ $dy/dx = (2(1-2x^2))/sqrt(1-x^2)$

Answer 2 · 2018-06-26T20:26:28

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

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

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

Apply the product rule etc to find:

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