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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 $\log _{10}$ (998). Given that $\log _{10} e=0.4343$.
✓

Answer

Let $f(x)=\log _{10} x=\frac{\log x}{\log 10}$
$
\therefore \quad f(x)=\left(\log _{10} e\right) \cdot \log x
$
Differentiate $w$. r.t. $x$.
$
f^{\prime}(x)=\frac{\log _{10} e}{x}=\frac{0.4343}{x}
$
Let $a=1000, h=-2$
For $x=a=1000$, from (I) we get
$
\begin{aligned}
f(a) & =f(1000)=\log _{10} 1000 \\
\therefore \quad f(a) & =3 \log _{10} 10=3
\end{aligned}
$
For $x=a=1000$, from (II) we get
$
\begin{gathered}
f^{\prime}(a)=f^{\prime}(1000)=\frac{0.4343}{1000} \\
\therefore \quad f^{\prime}(a)=0.0004343 \\
\text { We have, } f(a+h) \doteqdot f(a)+h f^{\prime}(a) \\
f[1000+(-2)] \doteqdot f(1000)+(-2) f^{\prime}(1000) \\
f(998) \doteqdot 3-(2)(0.0004343) \ldots \\
{[\text { From (III) and (IV) }]} \\
\quad \doteqdot 3-0.0008686 \\
f(998)=\log (998) \doteqdot 2.9991314
\end{gathered}
$

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