if y = log x base e, then find delta y when x= 3 and delta x= 0.03

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$$\begin{gathered} \mathrm{Given}, \\ \mathrm{y}\left(\mathrm{x}\right)={\log }_{\mathrm{e}}\mathrm{x} \\ \mathrm{Differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\log }_{\mathrm{e}}\mathrm{x}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}} \\ \mathrm{We} \mathrm{have} \mathrm{a} \mathrm{formula} \mathrm{for} \mathrm{approximation}, \\ \mathrm{\Delta y}\approx \mathrm{\Delta x}\frac{\mathrm{dy}}{\mathrm{dx}} \\ \mathrm{Given} \mathrm{that} \mathrm{x}=3 \mathrm{and} \mathrm{\Delta x}=0.03. \mathrm{Substituting}, \\ \mathrm{\Delta y}\approx \mathrm{\Delta x}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{\Delta x}·\left(\frac{1}{x}\right) \\ \mathrm{\Delta y}\approx 0.03\times \frac{1}{3} \\ \mathbf{\Delta y}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{01} \\ \mathrm{This} \mathrm{could} \mathrm{be} \mathrm{an} \mathrm{approximation} \mathrm{of} \mathrm{\Delta y}. \\ \mathrm{If} \mathrm{we} \mathrm{calculate}, \\ \mathrm{\Delta y}={\log }_{\mathrm{e}}\left(3.03\right)-{\log }_{\mathrm{e}}\left(3\right)\approx 0.00995033085317\approx 0.01 \\ \mathrm{This} \mathrm{seems} \mathrm{to} \mathrm{be} \mathrm{close} \mathrm{enough}. \end{gathered}$$

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