Question

Help with 27 please. thanks​

Answer 1

See Below.

Explanation:

We are given the function:

`\displaystyle y=√(\sin x)`

And we want to show that:

`\displaystyle 4y^3(d^2y)/(dx^2)+y^4+1=0`

Find the first derivative of y using the chain rule:

`\displaystyle (dy)/(dx) = (1)/(2√(\sin x))\cdot \cos x = (\cos x)/(2√(\sin x))`

And find the second derivative using the quotient and chain rules:

`\displaystyle \begin{aligned} (d^2y)/(dx^2) &= (1)/(2)\left(((\cos x)'(√(\sin x))-(\cos x)(√(\sin x))')/((√(\sin x))^2)\right) \\ \\ &=(1)/(2)\left((-\sin x√(\sin x) - \left(\cos x\right) \left ((\cos x)/(2√(\sin x))\right))/(\sin x)\right) \\ \\ & = (1)/(2)\left(( -\sin x(2\sin x) -\cos x(\cos x) )/(\sin x \left(2√(\sin x)\right) )\right) \\ \\ &= -(1)/(2) \left(\frac{2\sin^2 x + \cos^2 x}{2\sin^{{}^(3)\!/\! {}_(2)}x}\right)\end{aligned}`

Find y³:

`\displaystyle y^3 = \left((\sin x)^{{}^(1)\!/\!{}_(2)}\right) ^3= \sin^{{}^(3)\! / \! {}_(2) }x`

And find y⁴:

`\displaystyle y^4 = \left((\sin x)^{{}^(1)\!/\!{}_(2)}\right)^4 = \sin^2 x`

Substitute:

`\displaystyle 4\left( \sin^{{}^(3)\! / \! {}_(2) }x\right)\left(-(1)/(2)\left(\frac{2\sin ^2x + \cos ^2 x}{2\sin^{{}^(3)\!/ \! {}_(2)}x}\right)\right)+\left(\sin ^2 x\right) + 1= 0`

Simplify:

`-\left(2\sin^2 x + \cos^2 x\right) + \sin ^2 x + 1=0`

Distribute:

`-2\sin ^2 x - \cos^2 x + \sin ^2 x + 1=0`

Simplify:

`-\sin ^2 x - \cos^2 x + 1= 0`

Factor:

`-(\sin ^2 x + \cos^2 x ) + 1=0`

Pythagorean Identity:

`-(1)+1=0\stackrel{\checkmark}{=}0`

Q.E.D.