Question

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

Answer 1

`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))`