Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$
✓
Answer
Let, $\text{u}=\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ $\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\frac{\text{x}}{2}\Big)\Big]$ $\Rightarrow\text{u}=\frac{\pi}{4}-\frac{\text{x}}{2}$ Differentiating it with respect to x, $\frac{\text{du}}{\text{dx}}=0-\big(\frac{1}{2}\big)$ $\frac{\text{du}}{\text{dx}}=-\frac{1}{2}\ .....(\text{i})$ Let, $\text{v}=\sec^{-1}\text{x}$ Differentiating it with respect to x, $\frac{\text{dv}}{\text{dx}}=\frac{1}{\text{x}\sqrt{\text{x}^2-1}}\ .....\text{(ii)}$ Differentiating equation (i) by (ii), $\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{2}\times\frac{\text{x}\sqrt{\text{x}^2-1}}{1}$ $\frac{\text{du}}{\text{dv}}=\frac{-\text{x}\sqrt{\text{x}^2-1}}{2}$
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