Answer:
[tex]\qquad\qquad\boxed{ \sf{ \: \bf \:\dfrac{1}{ {(2.001)}^{2} } = 0.24975 \: }} \\ \\ [/tex]
Step-by-step explanation:
Let assume that
[tex]\sf \:f(x) = \dfrac{1}{ {x}^{2} }, \: \: where \: x = 2 \\ \\ [/tex]
So,
[tex]\sf \:f(x + \triangle x) = \dfrac{1}{ {(x + \triangle x)}^{2} }, \: \: where \: \triangle x = 0.001 \\ \\ [/tex]
We know, By definition of differentials, we have
[tex]\sf \:f(x + \triangle x) = f(x) + \dfrac{d}{dx}f(x) \times \triangle x \\ \\ [/tex]
[tex]\sf \:\dfrac{1}{ {(x + \triangle x)}^{2} } = \dfrac{1}{ {x}^{2} } + \dfrac{d}{dx}\left(\dfrac{1}{ {x}^{2} } \right) \times \triangle x \\ \\ [/tex]
[tex]\sf \:\dfrac{1}{ {(x + \triangle x)}^{2} } = \dfrac{1}{ {x}^{2} } - \dfrac{2}{ {x}^{3} } \times \triangle x \\ \\ [/tex]
On substituting the values x and [tex]\triangle [/tex]x, we get
[tex]\sf \:\dfrac{1}{ {(2+ 0.001)}^{2} } = \dfrac{1}{ {(2)}^{2} } - \dfrac{2}{ {(2)}^{3} } \times 0.001 \\ \\ [/tex]
[tex]\sf \:\dfrac{1}{ {(2.001)}^{2} } = \dfrac{1}{4} - \dfrac{1}{ 4 } \times 0.001 \\ \\ [/tex]
[tex]\sf \:\dfrac{1}{ {(2.001)}^{2} } = \dfrac{1}{4} - \dfrac{0.001}{4} \\ \\ [/tex]
[tex]\sf \:\dfrac{1}{ {(2.001)}^{2} } = \dfrac{1 - 0.001}{4} \\ \\ [/tex]
[tex]\sf \:\dfrac{1}{ {(2.001)}^{2} } = \dfrac{0.999}{4} \\ \\ [/tex]
[tex]\sf\implies \bf \:\dfrac{1}{ {(2.001)}^{2} } = 0.24975 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]