Linear Approximation and Differentials

In Fig. 7, the change in value of the function $\mathrm{f}$ near the point $(x, f(x))$ is $\Delta f=f(x+\Delta x)-f(x)$ and the change along the tangent line is $\mathrm{f}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x}$. If $\Delta \mathrm{x}$ is small, then we have used the approximation that $\Delta \mathrm{f} \approx \mathrm{f}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x}$. This leads to the definition of a new quantity, $\mathrm{df}$, called the differential of $\mathrm{f}$.

The differential of $f$ represents the change in $f$, as $x$ changes from $x$ to $x+d x$, along the tangent line to the graph of $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$. If we take $\mathrm{dx}$ to be the number $\Delta \mathrm{x}$, then the differential is an approximation of $\Delta f: \Delta f \approx f^{\prime}(x) \cdot \Delta x=f^{\prime}(x) \cdot d x=d f$.

Example 7: Determine the differential df of each of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-7 \mathrm{x}, \mathrm{g}(\mathrm{x})=\sin (\mathrm{x})$, and $\mathrm{h}(\mathrm{r})=\pi \mathrm{r}^{2}$. Solution: $\quad \mathrm{df}=\mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{d} \mathrm{x}=\left(3 \mathrm{x}^{2}-7\right) \mathrm{dx}, \mathrm{dg}=\mathrm{g}^{\prime}(\mathrm{x}) \cdot \mathrm{dx}=\cos (\mathrm{x}) \mathrm{dx}$, and $\mathrm{dh}=\mathrm{h}^{\prime}(\mathrm{r}) \mathrm{dr}=2 \pi \mathrm{r} \mathrm{dr}$.

Practice 9: Determine the differentials of $\mathrm{f}(\mathrm{x})=\ln (\mathrm{x}), \mathrm{u}=\sqrt{1-3 \mathrm{x}}$, and $\mathrm{r}=3 \cos (\theta)$.

We will do little with differentials for a while, but are used extensively in integral calculus.