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Applications of Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivatives3 Marks
Question
Find the approximate value of $\tan ^{-1}(1.001)$
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Answer

Let $f(y) = \tan^{-1}y$
Differentiating $f(y) w.r.t.y,$ we have
$\Rightarrow f ^{\prime}( y )=\frac{1}{1+y^2}$
$y=1.001=x+\Delta x$
Here,
$x = 1$
$\triangle x = 0.001$
Therefore, $f(x)=f(1)=\tan ^{-1}(1)=\frac{\pi}{4}$
Similarly, $f^{\prime}(x)=f^{\prime}(1)=\frac{1}{1+1^2}=\frac{1}{2}$
Now,
$f(y) = f( x + \triangle x ) = f(x) + \triangle x. f'(x) ...[ \because \triangle x <<< x ]$
$\tan ^{-1} y=\tan ^{-1}(x+\Delta x)=\tan ^{-1} x+\Delta x \cdot\left(\frac{1}{1+x^2}\right)$
$\therefore \tan^{-1}1.001 = \tan^{-1}( 1 + 1.001 ) = \tan^{-1}1 + (0.001). \tan^{-1}1$
$\Rightarrow \tan ^{-1} 1.001=\frac{\pi}{4}+0.001\left(\frac{1}{2}\right)$
$\Rightarrow \tan ^{-1} 1.001=\frac{\pi}{4}+0.0005 \approx 0.7855$
Hence the approxiate value of $\tan^{-1}0.001$ will be $0.7855.$

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