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Higher Order Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives4 Marks
Question
If $\text{y}=\sin(\sin\text{x})$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0$
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Answer

Given,
$\text{y} = \sin (\sin \text{x})\dots\text{ eq. } 1$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\sin\text{x})$
Using chain rule, we will differentiate the above expression:
Let $\text{t}=\sin\text{x}\Rightarrow\frac{\text{dt}}{\text{dx}}=\cos\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dy}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\cos\text{t}\cos\text{x}=\cos(\sin\text{x})\cos\text{x}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\cos\text{x}\frac{\text{d}}{\text{dx}}\cos(\sin\text{x})+\cos(\sin\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}$
Using chain rule again in the next step:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\cos\text{x}\cos\text{x}\sin(\sin\text{x})-\sin\text{x}\cos(\sin\text{x})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\cos^2\text{x}-\tan\text{x}\cos\text{x}\cos(\sin\text{x})$
$[$using eq. 1: $\text{y} = \sin (\sin \text{x})]$
And using eq. 2, we have:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\cos^2\text{x}-\tan\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}\cos^2\text{x}+\tan\text{x}\frac{\text{dy}}{\text{dx}}=0\dots\text{ proved.}$

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