Applications of Derivatives — Maths STD 12 Science — Question

Let $f(x)=\log _{10} x=\frac{\log _e x}{\log _e 10}$
$=\log _{10} e \log _e x=(0.4343) \log x$
$\Rightarrow f^{\prime}(x)=\frac{0.4343}{x}$
Now $,  x=998=1000-2=a-h$
Where $ a=1000$ and $ h=2$
$f(a)=\log _{10}(1000)=\log _{10}(10)^3$
$=3 \log _{10} 10=3$
$f^{\prime}(a)=f^{\prime}(1000)$
$=\frac{0.4343}{1000}=0.0004343$
Now, $f(a-h)=f(a)-h f^{\prime}(a)$
$=3-2(0.0004343)$
$=3-0.0008686=2.9991314$