How do you find the derivative of y = arcsin(1 - 2sin^2x)y=arcsin(1−2sin2x)?

2 Answers
Noah G
Oct 10, 2017

We can rewrite as

siny = 1 - 2sin^2xsiny=1−2sin2x

Which in turn can be rewritten as

siny = cos(2x)siny=cos(2x)

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

cosy(dy/dx) = -2sin(2x)cosy(dydx)=−2sin(2x)

dy/dx = (-2sin(2x))/cosydydx=−2sin(2x)cosy

We know that sin^2y + cos^2y = 1sin2y+cos2y=1 and so cosy = sqrt(1 -sin^2ycosy=√1−sin2y. 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)cosy=√1−(1−2sin2x)2=√1−(1−4sin2x+4sin4x)=√4sin2x−4sin4x. Accordingly,

dy/dx = (-2sin(2x))/sqrt(4sin^2x - 4sin^4x)dydx=−2sin(2x)√4sin2x−4sin4x

We can simplify as follows:

dy/dx = (-2(2sinxcosx))/sqrt(4sin^2x(1 - sin^2x))dydx=−2(2sinxcosx)√4sin2x(1−sin2x)

dy/dx= (-2(2sinxcosx))/(2sinxsqrt(1 - sin^2x))dydx=−2(2sinxcosx)2sinx√1−sin2x

dy/dx = (-2cosx)/sqrt(cos^2x)dydx=−2cosx√cos2x

dy/dx = (-2cosx)/|cosx|dydx=−2cosx|cosx|

Depending upon the value of xx, dy/dxdydx will either be positive or negative.

cosxcosx will negative whenever pi/2 ≤ x ≤ (3pi)/2π2≤x≤3π2.

Since the periodicity of cosxcosx is 2pi2π.

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!

sankarankalyanam
Oct 10, 2017

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)

Related questions
Impact of this question
1330 views around the world
You can reuse this answer
Creative Commons License