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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{y}=\Big[\log\Big(\text{x}+\sqrt{\text{x}^2+1}\Big)\Big]^2$ show that $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=2$
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Answer

Given:
$\text{y}=\Big[\log\Big(\text{x}+\sqrt{1+\text{x}}\Big)\Big]^2$
Differentiating w.r.t. x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}\big[\log\big(\text{x}+\sqrt{1+\text{x}^2}\big)\big]^2}{\text{dx}}$
Using formula (ii),
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\log\big(\text{x}+\sqrt{1+\text{x}}\big).\frac{1}{(\text{x}+\sqrt{1+\text{x}^2})}.\Big(1+\frac{2\text{x}}{2\sqrt{1+\text{x}^2}}\Big)$
Using formula (i),
$\Rightarrow\text{y}_1=\frac{2\log(\text{x}+\sqrt{1+\text{x}^2}}{\text{x}+\sqrt{1+\text{x}^2}}.\frac{\text{x}+\sqrt{1+\text{x}^2}}{\sqrt{1+\text{x}^2}}$
$\Rightarrow\text{y}_1=\frac{2\log(\text{x}\sqrt{1+\text{x}^2})}{\sqrt{1+\text{x}^2}}$
Squaring both sides:
$\text{(y}_1)^2=\frac{4}{1+\text{x}^2}[\log\big(\text{x}\sqrt{1+\text{x}^2}\big)$
Differentiating w.r.t. x,
$\Rightarrow(1+\text{x}^2)\text{y}_2\text{y}_1+2\text{x}(\text{y}_1)^2=4\text{y}_1$
Using formual (iii),
$\Rightarrow(1+\text{x}^2)\text{y}_2+\text{xy}_1=2$
Hence proved.

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