Question

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

Answer 1

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

Explanation:

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

`y=sin(2sin^-1(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) #