Question

How do you find the derivative of `y = arcsin(1 - 2sin^2x)`?

Answer 1

We can rewrite as

`siny = 1 - 2sin^2x`

Which in turn can be rewritten as

`siny = cos(2x)`

We now differentiate implicitly on the left side and using the chain rule on the right.

`cosy(dy/dx) = -2sin(2x)`

`dy/dx = (-2sin(2x))/cosy`

We know that `sin^2y + cos^2y = 1` and so `cosy = sqrt(1 -sin^2y`. From the original function, we know that `cosy = sqrt(1 - (1 - 2sin^2x)^2) = sqrt(1 - (1 - 4sin^2x + 4sin^4x)) = sqrt(4sin^2x - 4sin^4x)`. Accordingly,

`dy/dx = (-2sin(2x))/sqrt(4sin^2x - 4sin^4x)`

We can simplify as follows:

`dy/dx = (-2(2sinxcosx))/sqrt(4sin^2x(1 - sin^2x))`

`dy/dx= (-2(2sinxcosx))/(2sinxsqrt(1 - sin^2x))`

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

`dy/dx = (-2cosx)/|cosx|`

Depending upon the value of `x`, `dy/dx` will either be positive or negative.

`cosx` will negative whenever `pi/2 ≤ x ≤ (3pi)/2`.

Since the periodicity of `cosx` is `2pi`.

We can therefore define the derivative by the piecewise function

`dy/dx = {(2, "when " pi/2 + npi ≤ x ≤ pi +npi, x in RR, n in ZZ), (-2, "when " npi ≤ x ≤ pi/2 +npi, x in RR, n in ZZ):}`

Thanks to Douglas K for his guidance here!

Hopefully this helps!

Answer 2

`y’=(-2cosx)/sqrt(1+cos^2x)`

Explanation:

`y=sin^-1(1-2sin^2x)=sin^-1(cos^2x)`
Put `u=cos^2x`
`(du)/(dx)=-2cos x*sin x=-sin 2x`
`(dy)/(dx)=(d/dx)sin^-1u=(1/sqrt(1-u^2))(du)/(dx)`
`y’=(1/sqrt(1-cos^4x))*(-sin 2x)`
`y’=(-sin 2x)/(sqrt((1+cos^2x)(1-cos^2x)))`
`=(-2*cancel(sinx)*cosx)/(cancel(sinx)*sqrt(1+cos^2x))`
`y’=(-2cos x)/sqrt(1+cos^2x)`