Higher Order Derivatives — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives4 Marks
Question
If $\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
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Answer
Given, $\text{y}=(\sin^{-1}\text{x})^2\dots\text{ eq.1}$ To prove: $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$ 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 $\frac{\text{dy}}{\text{dx}}$ $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2$ Using chain rule we will differentiate the above expression: Let $\text{t}=\sin^{-1}\text{x}$ $\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\sqrt{(1-\text{x}^2)}}$$[$using formula for derivative of $\sin^{-1}\text{x}]$ And y = t2 $\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$ $\frac{\text{dy}}{\text{dx}}=2\text{t}\frac{1}{\sqrt{(1-\text{x}^2)}}=2\sin^{-1}\text{x}\frac{1}{\sqrt{(1-\text{x}^2)}}\dots\text{ eq. 2}$ Again differentiating with respect to x applying product rule: $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\frac{2}{\sqrt{(1-\text{x}^2)}}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}$ $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2\sin^{-1}\text{x}}{2(1-\text{x}^2)\sqrt{1-\text{x}^2}}(-2\text{x})+\frac{2}{(1-\text{x}^2)}$$\bigg[\text{using }\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n}-1}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}=\frac{1}{\sqrt{(1-\text{x}^2)}}\bigg]$ $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}+\frac{2}{(1-\text{x}^2)}$ $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{2\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$ Using eq. 2: $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{\text{dy}}{\text{dx}}$ $\therefore(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0\dots\text{ proved.}$
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