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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate w.r.t. x the function in Exercise:
$(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})},\ \frac{\pi}{4}<\text{x}<\frac{3\pi}{4}$
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Answer

Let $\text{y}=(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})}$
Tanking logarithm on both the sides, we obtain
$\log\text{y}=\log\big[(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})}\big]$
$\Rightarrow\ \log\text{y}=(\sin\text{x}-\cos\text{x)}.\log(\sin\text{x}-\cos\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[(\sin\text{x}-\cos\text{x})\log(\sin\text{x}-\cos\text{x})]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x}).\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})+(\sin\text{x}-\cos\text{x}).\frac{\text{d}}{\text{dx}}\log(\sin\text{x}-\cos\text{x})$
$\Rightarrow\ \ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x}).(\cos\text{x}+\sin\text{x})+(\sin\text{x}-\cos\text{x}).\frac{1}{(\sin\text{x}-\cos\text{x})}.\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x)}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})}[(\cos\text{x}+\sin\text{x}).\log(\sin\text{x}-\cos\text{x})+(\cos\text{x}+\sin\text{x})]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})^{(\sin\text{x}-\cos\text{x})}(\cos\text{x}+\sin\text{x})[1+\log(\sin\text{x}-\cos\text{x})]$

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