How do you find the derivative of #y = arcsin(1 - 2sin^2x)#?
2 Answers
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
#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
Since the periodicity of
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!
Explanation:
Put