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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following w.r.t. x:
$\cos^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big),\frac{-\pi}{4}<\text{x}<\frac{\pi}{4}$
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Answer

Let $\text{y}=\cos^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\cos^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)$
$=\frac{-1}{\sqrt{1-\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)^2}}\cdot\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)$
$\Big[\because\ \frac{\text{d}}{\text{dx}}(\cos\text{x})=-\frac{1}{\sqrt{1-\text{x}^2}}\Big]$
$=\frac{-1}{\sqrt{4-\frac{(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cdot\cos\text{x})}{2}}}\cdot\frac{1}{\sqrt{2}}(\cos\text{x}-\sin\text{x})$
$=\frac{-1\cdot\sqrt{2}}{\sqrt{1-\sin2\text{x}}}\cdot\frac{1}{\sqrt{2}}(\cos\text{x}-\sin\text{x})$
$\big[\because1-\sin2\text{x}=(\cos\text{x}-\sin\text{x})^2=\cos^2\text{x}+\sin^2\text{x}-2\sin\text{x}\cos\text{x}\big]$
$=\frac{-1(\cos\text{x}-\sin\text{x})}{(\cos\text{x}-\sin\text{x})}=-1$

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