Question

Hi, help with question 18 please. thanks​

Answer 1

See Below.

Explanation:

We are given the equation:

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

And we want to prove that:

`\displaystyle 2y(d^2y)/(dx^2) + 2\left((dy)/(dx)\right) ^2 + y^2 = 1`

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

`\displaystyle 2y (dy)/(dx) = \cos x`

Divide both sides by 2y:

`\displaystyle (dy)/(dx) = (\cos x)/(2y)`

Find the second derivative using the quotient rule:

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

Substitute:

`\displaystyle 2y\left(-(2y^2\sin x+\cos ^2 x)/(4y^3)\right) + 2\left((\cos x)/(2y)\right)^2 +y^2 = 1`

Simplify:

`\displaystyle (-2y^2\sin x-\cos ^2x)/(2y^2) + (\cos ^2 x)/(2y^2) + y^2 = 1`

Combine fractions:

`\displaystyle (\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right))/(2y^2) + y^2 = 1`

Simplify:

`\displaystyle (-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.