Explain Derivative???

1-sin^2 y = cos^2 y

sqrt (cos^2 y) = cos y

dy/dx = 2x / cos y

this is a separable, first-order differential equation. ur supposed to cross-multiply and integrate both sides.

cos y dy = 2x dx

integral (cos y dy) = integral (2x dx)

sin y = x^2 +C

y = arcsin (x^2 +C).

this is called the general solution for the differential equation.

now if we had some intial value like y(pi) =0, we could find the C.

a differential equation without C is called a ;particular solution.

dy/dx =2x/√(1-sin^2(y)) = 2x / √(cos^2(y)) = 2x / | cos(y) |

If cos(y) = √(1-x^4) then

dy/dx = 2x/√(1-x^4)

Otherwise I have no idea where 2x/√(1-x^4) came from.

Edit:

That's just a simple substitution then.

Maybe this notation is confusing you "sin^2(y)"

sin^2(y) is the same thing as (sin(y))^2

Since sin(y) = x^2 you can just substitute this into the original equation

dy/dx = 2x / √(1-(sin(x))^2)

= 2x / √(1 - (sin(x))^2)

= 2x / √(1 - (x^2)^2)

= 2x / √(1 - x^4)

dy/dx = 2x / √(1 − sin²(y) )

√(1 − sin²(y) ) dy = 2x dx

√(cos²(y) ) dy = 2x dx

cos(y) dy = 2x dx

∫ cos(y) dy = ∫ 2x dx

sin(y) = x² + C

y = arcsin(x² + C)

I dont know what youre asking