Question

Hi, help with question 18 please. thanks​

Answer 1

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

$\displaystyle y^2 = 1 + \sin x$

And we want to prove that:

$\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1$

Find the first derivative by taking the derivative of both sides with respect to x:

$\displaystyle 2y \frac{dy}{dx} = \cos x$

Divide both sides by 2y:

$\displaystyle \frac{dy}{dx} = \frac{\cos x}{2y}$

Find the second derivative using the quotient rule:

$\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\ &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}$

Substitute:

$\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right) + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1$

Simplify:

$\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1$

Combine fractions:

$\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1$

Simplify:

$\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1$

Cancel:

$\displaystyle -\sin x + y^2 = 1$

Substitute:

$-\sin x + \left( 1 + \sin x\right) =1$

Simplify. Hence:

$1\stackrel{\checkmark}{=}1$

Q.E.D.