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Differentiation — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
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Answer

Let $\text{y}=(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\cos\text{x})^\text{x}}+\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log(\cos\text{x})}+\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\cos\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}\big)$
$=\text{e}^{\log\cos\text{x}}\times\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{x})+\text{e}^{\frac{1}{\text{x}}\log\sin}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\log\sin\text{x}\big)$
$=\text{e}^{\log(\cos\text{x})^\text{x}}\times\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\times\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}\times\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}+\log\cos\text{x}(1)\Big] \\ +(\sin)^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\cos\text{x}}\Big)(-\sin\text{x})+\log\cos\text{x}\Big] \\ +(\sin\text{x})^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}(\cos\text{x})-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
$=(\cos\text{x})^\text{x}\big[\log\cos\text{x}-\text{x}\tan\text{x}\big](\sin\text{x})^\frac{1}{\text{x}} \\ \Big[\frac{\cot\text{x}}{\text{x}}-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$

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