Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
✓
Answer
Here,
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}+\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}$
$\big[\text{Since},\log_\text{e}\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\cot\text{x}\log\tan\text{x}}+\text{e}^{\tan\text{x}\log(\cot\text{x})}$
Differentiating it with respect to x using rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cot\text{x}\log\tan\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\cot\text{x}}\big)$
$=\text{e}^{\cot\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\cot\text{x}\log\tan\text{x})+\text{e}^{\tan\text{x}\log\cot\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\cot\text{x})$
$=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}\Big[\cot\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{x}+\log\tan\text{x}\frac{\text{d}}{\text{dx}}\cot\text{x}\Big] \\ +\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}\Big[ \tan\text{x}\frac{\text{d}}{\text{dx}} \log\cot\text{x}+\log\cot\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[\cot\text{x}\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}(-\text{cosec}^2\text{x})\Big] \\ +(\cot\text{x})^{\tan\text{x}}\Big[\tan\text{x}\big(\frac{1}{\cot\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cot\text{x})+\log\cot\text{x}\big(\sec^2\text{x}\big)\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[(1)\big(\sec^2\text{x}\big)-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[(1)\big(-\text{cosec}^2\text{x}\big)+\sec^2\text{x}\log\cot\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\cot\text{x}}\Big[\sec^{2\text{x}}-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[\sec^2\text{x}\log\cot\text{x}-\text{cosec}^2\text{x}\Big]$
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}+\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}$
$\big[\text{Since},\log_\text{e}\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\cot\text{x}\log\tan\text{x}}+\text{e}^{\tan\text{x}\log(\cot\text{x})}$
Differentiating it with respect to x using rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cot\text{x}\log\tan\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\cot\text{x}}\big)$
$=\text{e}^{\cot\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\cot\text{x}\log\tan\text{x})+\text{e}^{\tan\text{x}\log\cot\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\cot\text{x})$
$=\text{e}^{\log(\tan\text{x})^{\cot\text{x}}}\Big[\cot\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{x}+\log\tan\text{x}\frac{\text{d}}{\text{dx}}\cot\text{x}\Big] \\ +\text{e}^{\log(\cot\text{x})^{\tan\text{x}}}\Big[ \tan\text{x}\frac{\text{d}}{\text{dx}} \log\cot\text{x}+\log\cot\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[\cot\text{x}\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}(-\text{cosec}^2\text{x})\Big] \\ +(\cot\text{x})^{\tan\text{x}}\Big[\tan\text{x}\big(\frac{1}{\cot\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cot\text{x})+\log\cot\text{x}\big(\sec^2\text{x}\big)\Big]$
$=(\tan\text{x})^{\cot\text{x}}\Big[(1)\big(\sec^2\text{x}\big)-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[(1)\big(-\text{cosec}^2\text{x}\big)+\sec^2\text{x}\log\cot\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\cot\text{x}}\Big[\sec^{2\text{x}}-\text{cosec}^2\text{x}\log\tan\text{x}\Big] \\ +(\cot)^{\tan\text{x}}\Big[\sec^2\text{x}\log\cot\text{x}-\text{cosec}^2\text{x}\Big]$
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