Answer:
See Below.
Step-by-step explanation:
We are given the equation:
[tex]\displaystyle y^2 = 1 + \sin x[/tex]
And we want to prove that:
[tex]\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1[/tex]
Find the first derivative by taking the derivative of both sides with respect to x:
[tex]\displaystyle 2y \frac{dy}{dx} = \cos x[/tex]
Divide both sides by 2y:
[tex]\displaystyle \frac{dy}{dx} = \frac{\cos x}{2y}[/tex]
Find the second derivative using the quotient rule:
[tex]\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\ &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}[/tex]
Substitute:
[tex]\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right) + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1[/tex]
Simplify:
[tex]\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1[/tex]
Combine fractions:
[tex]\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1[/tex]
Simplify:
[tex]\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1[/tex]
Cancel:
[tex]\displaystyle -\sin x + y^2 = 1[/tex]
Substitute:
[tex]-\sin x + \left( 1 + \sin x\right) =1[/tex]
Simplify. Hence:
[tex]1\stackrel{\checkmark}{=}1[/tex]
Q.E.D.
