How do you estimate the quantity using linear approximation and find the error using a calculator of $\dfrac{1}{\sqrt{95}}-\dfrac{1}{\sqrt{98}}$ ?

Hint: For answering this question we will assume a function $f\left( x \right)=\dfrac{1}{\sqrt{x}}$ and find the slope${{f}^{'}}\left( x \right)$ of the following curve. We will use the equation of the curve and find the approximate value and compare with the calculated value.

Complete step by step solution:
Now considering from the question we have been asked to estimate the quantity using linear approximation and find the error using a calculator of $\dfrac{1}{\sqrt{95}}-\dfrac{1}{\sqrt{98}}$ .
We will assume a function $f\left( x \right)=\dfrac{1}{\sqrt{x}}$ with slope $\dfrac{d}{dx}f\left( x \right)=\dfrac{-1}{2}{{x}^{\dfrac{-3}{2}}}$ .
Let us consider $x=100$ then we will have $f\left( 100 \right)=\dfrac{1}{10}$ and slope
$\begin{align}
  & \dfrac{d}{dx}f\left( 100 \right)=\dfrac{-1}{2}{{\left( 100 \right)}^{\dfrac{-3}{2}}} \\
 & \Rightarrow \dfrac{d}{dx}f\left( 100 \right)=\dfrac{-1}{2}{{\left( 10 \right)}^{-3}} \\
\end{align}$ .
Now we will try to get the equation of the curve. The equation of the curve will be given as
$\begin{align}
  & \left( y-f\left( {{x}_{1}} \right) \right)={{f}^{'}}\left( {{x}_{1}} \right)\left( x-{{x}_{1}} \right) \\
 & \Rightarrow \left( y-\dfrac{1}{10} \right)=\dfrac{-1}{2}{{\left( 10 \right)}^{-3}}\left( x-100 \right) \\
 & \Rightarrow y=\dfrac{-1}{2000}x+\dfrac{3}{20} \\
\end{align}$ .
Now we need to get the value of
$\begin{align}
  & f\left( 95 \right)-f\left( 98 \right)=\dfrac{-1}{2000}\left( 95 \right)+\dfrac{3}{20}-\left( \dfrac{-1}{2000}\left( 98 \right)+\dfrac{3}{20} \right) \\
 & \Rightarrow \dfrac{-1}{2000}\left( 95 \right)+\dfrac{3}{20}+\dfrac{1}{2000}\left( 98 \right)-\dfrac{3}{20}=\dfrac{1}{2000}\left( 98-95 \right) \\
 & \Rightarrow \dfrac{3}{2000}=0.00150 \\
\end{align}$.
Hence we can say that the estimated value is $0.00150$ and the calculator value is given as $0.00158$ .
The error percentage will be $%error=\dfrac{0.00150-0.00158}{0.00158}\Rightarrow 5%$ .
Hence we can say that the percentage error is $5%$ .
Therefore we can conclude that the estimated value of the given quantity $\dfrac{1}{\sqrt{95}}-\dfrac{1}{\sqrt{98}}$ using linear approximation is $0.00150$ and the value found with calculator is $0.00158$ and the error is $5%$.

Note: While answering this question we should be sure with our concept that we apply and the calculations we make. This is a question related to errors and approximations chapter. We can find the value of $\dfrac{1}{\sqrt{95}}$ and $\dfrac{1}{\sqrt{98}}$ as given
$\begin{align}
  & \dfrac{1}{\sqrt{95}}=\dfrac{-1}{2000}\left( 95 \right)+\dfrac{3}{20} \\
 & \Rightarrow \dfrac{1}{\sqrt{95}}=\dfrac{-95}{2000}+\dfrac{3}{20} \\
 & \Rightarrow \dfrac{1}{\sqrt{95}}=-0.0475+0.15 \\
 & \Rightarrow \dfrac{1}{\sqrt{95}}=-0.1025 \\
\end{align}$
and similarly
$\begin{align}
  & \dfrac{1}{\sqrt{98}}=\dfrac{-1}{2000}\left( 98 \right)+\dfrac{3}{20} \\
 & \Rightarrow \dfrac{1}{\sqrt{98}}=\dfrac{-98}{2000}+\dfrac{3}{20} \\
 & \Rightarrow \dfrac{1}{\sqrt{98}}=-0.049+0.15 \\
 & \Rightarrow \dfrac{1}{\sqrt{98}}=-0.101 \\
\end{align}$.