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Higher Order Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives3 Marks
Question
If $\text{y}=(\tan^{-1}\text{x})^2$ then prove that $(1+\text{x}^2)\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
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Answer

Here,
$\text{y}=(\tan^{-1}\text{x})^2$
Differentiating w.r.t.x, we get
$\text{y}_1=\frac{2\tan^{-1}\text{x}}{1+\text{x}^2}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{2-4\text{x}\tan^{-1}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\text{y}_2=\frac{2}{(1+\text{x}^2)^2}-\frac{2\tan^{-1}\text{x}\times2\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\text{y}_2=\frac{2}{(1+\text{x}^2)^2}-\frac{2\text{xy}_1}{(1+\text{x}^2)^2}$
$\Rightarrow(1+\text{x}^2)^2\text{y}_2=2-2\text{x}(1+\text{x}^2)\text{y}_1$
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_2=2$
Hence proved

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