Applications of Derivatives — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivatives3 Marks
Question
Find the approximate value of $\tan ^{-1}(0.99)$, Given that $\pi \doteqdot 3.1416$.
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Answer
Let $f(x)=\tan ^{-1} x$ Differentiate w.r.t. $x$. $f^{\prime}(x)=\frac{1}{1+x^2}$ Let $a=1, h=-0.01$ For $x=a=1$, from (I) we get $f(a)=f(1)=\tan ^{-1}(1)=\frac{\pi}{4}$ For $x=a=1$, from (II) we get $f^{\prime}(a)=f^{\prime}(1)=\frac{1}{1+1^2}=0.5$ We have, $f(a+h) \doteqdot f(a)+h f^{\prime}(a)$ $f[(1)+(-0.01)] \doteqdot f(1)+(-0.01) \cdot f^{\prime}(1)$ $f(0.99) \doteqdot \frac{\pi}{4}-(0.01) \cdot(0.5) \ldots[$ From (III) and (IV)] $\doteqdot \frac{\pi}{4}-0.005$ $\doteqdot \frac{3.1416}{4}-0.005$ $\doteqdot 0.7854-0.005=0.7804$ $\therefore f(0.99)=\tan ^{-1}(0.99) \doteqdot 0.7804$
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