Question
If $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}},\frac{\pi}{4}<\text{x}<\frac{3\pi}{4},$ find $\frac{\text{dy}}{\text{dx}}$

Answer

We have, $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\log\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x}) \\ +(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}\log(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})(\sin\text{x}-\cos\text{x}) \\ +\frac{(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})}\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})\log(\sin\text{x}-\cos\text{x})+(\cos\text{x}+\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})[1+\log(\sin\text{x}-\cos\text{x})]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})^{(\sin\text{x}-\cos\text{x})} \\ \big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$

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