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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\cos\theta,\text{y}=\sin^3$ prove that $\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=3\sin^2\theta(5\cos^2\theta-1)$
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Answer

Here$\text{x}=\cos\theta,\text{y}=\sin^3$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=3\sin^2\theta\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\sin^2\theta\cos\theta}{-\sin\theta}=-3\sin\theta\cos\theta$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-3\cos^2\theta+3\sin^2\theta)\frac{\text{d}\theta}{\text{dx}}\frac{)-3\cos^2\theta+3\sin^2\theta)}{-\sin\theta}$
Now,
$\text{LHS}=\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$=\sin^3\theta\times\frac{(-3\cos^2\theta+3\sin^2\theta)}{\sin\theta}+(-3\sin\theta\cos\theta)^2$
$=3\sin^2\theta\cos^2\theta-3\sin^4\theta+9\sin^2\theta\cos^2\theta$
$=12\sin^2\theta\cos^2\theta-3\sin^4\theta$
$=3\sin^2\theta(4\cos^2\theta-\sin^2\theta)$
$=3\sin^2\theta(5\cos^2\theta-1)$
$=\text{RHS}$

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