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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\text{x}^{(\sin\text{x}-\cos\text{x})}+\frac{\text{x}^2-1}{\text{x}^2+1}$
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Answer

Let $\text{y}=\text{x}^{(\sin\text{x}-\cos\text{x})}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\text{y}=\text{e}^{\log\text{x}^{\sin\text{c}-\cos\text{x}}}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\text{y}=\text{e}^{(\sin\text{c}-\cos\text{x})\log\text{x}}+\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and quotient rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{e}^{(\sin\text{x}-\cos\text{x})\log\text{x}}\Big]+\frac{\text{d}}{\text{dx}}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}\Big]$
$=\text{e}^{(\sin\text{x}-\cos\text{x})\log\text{x}}+\frac{\text{d}}{\text{dx}}\big\{(\sin\text{x}-\cos\text{x})\log\text{x}\big\} \\ +\bigg[\frac{(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)-(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)}{(\text{x}^2+1)^2}\bigg]$
$=\text{e}^{\log\text{x}^{(\sin\text{x}-\cos\text{x})}}\Big[(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})\Big] \\+\Big[\frac{(\text{x}^2+1)(2\text{x})-(\text{x}^2-1)(2\text{x})}{(\text{x}^2+1)^2}\Big]$
$=\text{e}^{(\sin\text{x}-\cos\text{x})}\Big[(\sin\text{x}-\cos\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(\sin\text{x}+\cos\text{x})\Big] \\ +\Big[\frac{2\text{x}^3+2\text{x}-2\text{x}^3+2\text{x}}{(\text{x}^2+1)^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}-\cos\text{x}}\Big[\frac{(\sin\text{x}-\cos\text{x})}{\text{x}}+\log\text{x}(\sin\text{x}+\cos\text{x})\Big]+\frac{4\text{x}}{(\text{x}^2+1)^2}$

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