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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{y}\log(1+\cos\text{x}),$ prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^\text{y}}{\text{dx}^2}.\frac{\text{d}\text{y}}{\text{dx}}=0$
✓

Answer

$\text{y}\log(1+\cos\text{x}),$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\cos\text{x}}\times-\sin\text{x}=\frac{-\sin\text{x}}{1+\cos\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{(1+\cos\text{x})\cos\text{x}-\sin(-\sin\text{x})}{(1+\cos\text{x})^2}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]=-\Big[\frac{1+\cos\text{x}}{(1+\cos\text{x})^2}\Big]=\frac{-1}{1+\cos\text{x}}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^3}=-\Big(\frac{+1}{(1+\cos\text{x})^2}\times+\sin\text{x}\Big)\\=-\Big(\frac{-\sin\text{x}}{1+\cos\text{x}}\Big)\times\Big(\frac{-1}{1+\cos\text{x}}\Big)=-\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}}{\text{dx}^2}.\frac{\text{dy}}{\text{dx}}=0$

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