If $y = \sin(\sin x)$, prove that $\dfrac{d^2 y}{dx^2} + an x\, \dfrac{dy}{dx} + y\cos^2 x = 0$.

Question

If $y = \sin(\sin x)$, prove that $\dfrac{d^2 y}{dx^2} + 	an x\, \dfrac{dy}{dx} + y\cos^2 x = 0$.

Accepted Answer

>

> [!TLDR]
> Substituting $y'$ and $y''$ makes every term cancel, proving the identity.
>
> **Concept.** Repeated use of the chain and product rules.
>
> **Why this method.** Compute the two derivatives explicitly and plug into the left side.
>
> **Working.** $y=\sin(\sin x)$.
> $$\frac{dy}{dx}=\cos(\sin x)\cdot\cos x.$$
> $$\frac{d^2y}{dx^2}=-\sin(\sin x)\cos x\cdot\cos x+\cos(\sin x)\cdot(-\sin x)=-\sin(\sin x)\cos^2x-\sin x\cos(\sin x).$$
> Now
> $$	an x\,\frac{dy}{dx}=	an x\cdot\cos(\sin x)\cos x=\sin x\cos(\sin x),$$ …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.If $y = \sin(\sin x)$ , prove that $\dfrac{d^2 y}{dx^2} + \tan x\, \dfrac{dy}{dx} + y\cos^2 x = 0$ .

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Q.If y=sin⁡(sin⁡x)y = \sin(\sin x)y=sin(sinx), prove that d2ydx2+tan⁡x dydx+ycos⁡2x=0\dfrac{d^2 y}{dx^2} + \tan x\, \dfrac{dy}{dx} + y\cos^2 x = 0dx2d2y​+tanxdxdy​+ycos2x=0.

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Q.If $y = \sin(\sin x)$ , prove that $\dfrac{d^2 y}{dx^2} + \tan x\, \dfrac{dy}{dx} + y\cos^2 x = 0$ .

Unlock everything free for 14 days