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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate $(\cos\text{x})^{\sin\text{x}}$ with respect to $(\sin\text{x})^{\cos\text{x}}$
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Answer

Let, $\text{u} = (\cos)^{\sin\text{x}}$
Taking log on both sides,
$\log\text{u} = \log(\cos\text{x})^{\sin \text{x}}$
$\Rightarrow \log\text{u} = \sin \text{x}\log(\cos\text{x})$
Differentiating it with respect to x using rule,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin \text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log \cos \text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
[using product rule]
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin \text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}(\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}[(\tan\text{x})\times(-\sin\text{x})+\log\log\text{x}(\cos\text{x})]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\cos\text{x})^{\sin\text{x}}[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}]\ .....\text{(i)} $
Let, $\text{v = }(\sin\text{x})^{\cos\text{x}}$
Taking log on both sides,
$\log\text{v}=\log(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{v}=\cos\text{x}\log(\sin\text{x})$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x}) $
[using product rule]
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\big(\frac{1}{\sin\text{x}}\big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}[\cot\text{x}(\cos \text{x})-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cot \text{x} (\cos \text{x})-\sin \text{x}\log\sin\text{x}]\ ...(\text{ii})$
Dividing equation (i) by (ii)
$\therefore\frac{\text{du}}{\text{dv}}=\frac{(\cos\text{x})^{\sin\text{x}}[\cos \text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x ]}}{(\sin\text{x})^{\cot\text{x}}[\cot \text{x}(\cos\text{x})-\sin\text{x}\log\sin\text{x}]}$

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