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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find the second order derivatives of the following functions:
$\text{y}=\text{x}.\cos\text{x}$
✓

Answer

let $\text{y}=\text{x}.\cos\text{x}$
Then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}.\cos\text{x})=\cos\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$=\cos.1+\text{x}(-\sin\text{x})=\cos\text{x}-\text{x}\sin\text{x}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}[\cos\text{x}-\text{x}\sin\text{x}]=\frac{\text{d}}{\text{dx}}(\cos\text{x})-\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x})$
$=-\sin\text{x}-\Big[\sin\text{x}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=-\sin\text{x}=(\sin\text{x}+\text{x}\cos\text{x})$

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