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Prove the following.
If y = cos(sin x), show that: (d²y)/(dx²) + tan x (dy)/(dx) + y cos² x = 0
If $y = \cos(\sin x)$, show that:
$\frac{d^2y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0$
Given: $y = \cos(\sin x)$
To show: $\frac{d^2y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0$
Answer
Answer
Differentiate twice and substitute.
- Step 1: $y = \cos(\sin x)$.
- Step 2: $\frac{dy}{dx} = -\sin(\sin x)\cdot\cos x$, by the chain rule.
- Step 3: $\frac{d^2y}{dx^2} = -\cos(\sin x)\cdot\cos x\cdot\cos x - \sin(\sin x)\cdot(-\sin x) = -\cos(\sin x)\cos^2 x + \sin x\sin(\sin x)$, by the product and chain rules.
- Step 4: $\tan x\,\frac{dy}{dx} = -\frac{\sin x}{\cos x}\sin(\sin x)\cos x = -\sin x\sin(\sin x)$.
- Step 5: $\frac{d^2y}{dx^2} + \tan x\,\frac{dy}{dx} = -\cos(\sin x)\cos^2 x + \sin x\sin(\sin x) - \sin x\sin(\sin x) = -y\cos^2 x$, since $y = \cos(\sin x)$.
- Step 6: Hence $\frac{d^2y}{dx^2} + \tan x\,\frac{dy}{dx} + y\cos^2 x = 0$.