Question

Help with 27 please. thanks​

Answer 1

Answer:

See Below.

Step-by-step explanation:

We are given the function:

$\displaystyle y=\sqrt{\sin x}$

And we want to show that:

$\displaystyle 4y^3\frac{d^2y}{dx^2}+y^4+1=0$

Find the first derivative of y using the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x}}\cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}}$

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

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

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:

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

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.