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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following w.r.t. x:
$\tan^{-1}\Big(\frac{\text{a}\cos\text{x}-\text{b}\sin\text{x}}{\text{b}\cos\text{x}+\text{a}\sin\text{x}}\Big),-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}\text{ and }\frac{\text{a}}{\text{b}}\tan\text{x}>-1$
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Answer

Let $\text{y}=\tan^{-1}\Big(\frac{\text{a}\cos\text{x}-\text{b}\sin\text{x}}{\text{b}\cos\text{x}+\text{a}\sin\text{x}}\Big)$
$=\tan^{-1}\Bigg[\frac{\frac{\text{a}\cos\text{x}}{\text{b}\cos\text{x}}-\frac{\text{b}\sin\text{x}}{\text{b}\cos\text{x}}}{\frac{\text{b}\cos\text{x}}{\text{b}\cos\text{x}}+\frac{\text{a}\sin\text{x}}{\text{b}\cos\text{x}}}\Bigg]$
$=\tan^{-1}\Bigg[\frac{\frac{\text{a}}{\text{b}}-\tan\text{x}}{1+\frac{\text{a}}{\text{b}}\tan\text{x}}\Bigg]$
$=\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\tan\text{x}$ $\bigg[\because\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\bigg]$
$\therefore\ \text{y}=\tan^{-1}\frac{\text{a}}{\text{b}}-\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)-\frac{\text{d}}{\text{dx}}(\text{x})$
$=0-1\bigg[\because\ \frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}}{\text{b}}\Big)=0\bigg]$
$=-1$

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