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

2 Answers
Jun 26, 2018

dy/dx = (2(1-2x^2))/sqrt(1-x^2) dydx=2(12x2)1x2

Explanation:

Let t=arcsinxt=arcsinx, which is defined for x in (-pi/2,pi/2)x(π2,π2)

Then:

sint = xsint=x

and

cost = sqrt(1-x^2)cost=1x2

because for x in (-pi/2,pi/2)x(π2,π2) the cosine is positive.

So:

y= sin(2arcsinx) = sin2t = 2sintcost = 2xsqrt(1-x^2)y=sin(2arcsinx)=sin2t=2sintcost=2x1x2

and using the product rule:

dy/dx = 2xd/dx(sqrt(1-x^2)) +2sqrt(1-x^2)dydx=2xddx(1x2)+21x2

dy/dx = 2x(-(2x)/(2sqrt(1-x^2))) +2sqrt(1-x^2)dydx=2x(2x21x2)+21x2

dy/dx = -(2x^2)/sqrt(1-x^2) +2sqrt(1-x^2)dydx=2x21x2+21x2

dy/dx = (-2x^2+2-2x^2)/sqrt(1-x^2) dydx=2x2+22x21x2

dy/dx = (2(1-2x^2))/sqrt(1-x^2) dydx=2(12x2)1x2

Jun 26, 2018

y=sin(2sin^-1(x))y=sin(2sin1(x))

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

  • sin alpha = sin 2 beta = 2 sinbeta cosbeta

  • {(sinalpha = y),(sinbeta = x),(cosbeta= sqrt(1 - sin^2 beta) = sqrt(1-x^2)):}

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)