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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{y}=\text{e}^{\tan^{-1}\text{x}},$ prove that $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
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Answer

Given,
$\text{y}=\text{e}^{\tan^{-1}\text{x}}\dots\text{ eq. }1$
To prove: $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
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, let’s first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{e}^{\tan^{-1}\text{x}}$
Using chain rule we will differentiate the above expression:
Let $\text{t}=\tan^{-1}\text{x}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{1+\text{x}^2}\Big[\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}=\frac{1}{1+\text{x}^2}\Big]$
And $y = e^t$​​​​​​​
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{t}}\frac{1}{1+\text{x}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{\tan^{-1}\text{x}}\Big(\frac{1}{1+\text{x}}\Big)+\frac{1}{1+\text{x}^2}\frac{\text{d}}{\text{dx}}\text{e}^{\tan^{-1}\text{x}}$
Using chain rule we will differentiate the above expression:
$\frac{\text{d}^2\text{y}}{\text{dx}}=\Big(\frac{\text{e}^{\tan^{-1}\text{x}}}{(1+\text{x}^2)^2}\Big)\frac{2\text{xe}^{\tan^{-1}\text{x}}}{(1+\text{x}^2)^2}$ $\Big[\text{using eq. 2;}\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{x}-1}\text{ and }\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}=\frac{1}{1+\text{x}^2}\Big]$
$(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}-\frac{2\text{xe}^{\tan^{-1}\text{x}}}{1+\text{x}^2}$
$(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}(1-2\text{x})$
Using equation 2:
$(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}(1-\text{2x})$
$\therefore(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0\dots\text{ proved.}$

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