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) #