How do you find #(dy)/(dx)# given #cos(2y)=sqrt(1-x^2)#?

1 Answer
Dec 13, 2016

# dy/dx = x/(2 sin(2y)sqrt(1 - x^2)) #

or equivalently:

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

Explanation:

# \ \ \ \ \ cos(2y)=sqrt(1-x^2) #
# :. cos(2y)=(1-x^2)^(1/2) #

Differentiating implicitly and applying the chain rule we get:
# -sin(2y)(2 dy/dx) = 1/2(1 - x^2)^(-1/2) (-2x) #
# 2 dy/dx sin(2y) = x/sqrt(1 - x^2) #

So we can rearrange to get;
# dy/dx = x/(2 sin(2y)sqrt(1 - x^2)) #

We can also get an explicit expression should we need it;

Using #sin^2A+cos^2A-=1# we have:

# sin^2 2y+cos^2 2y=1 #
# :. sin^2 2y+(1-x^2)=1 #
# :. sin^2 2y=x^2 #
# :. sin 2y=x #

So the earlier solution can be written as:

# \ \ \ \ \ dy/dx = x/(2xsqrt(1 - x^2)) #
# :. dy/dx = 1/(2sqrt(1 - x^2)) #