Differential of a Function

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/differential-of-a-function/ Fetched from algebrica.org post 6161; source modified 2025-05-17T16:55:19.

Consider $f(x)$ a differentiable function on the interval $[a,b]$. Since the function is differentiable, it is also continuous on the given interval. Let us consider two points $x$ and $x + \Delta x \in [a,b]$.

It is defined the differential of a function $f(x)$, relative to the point $x$ and the increment $\Delta x$, as the product of the derivative of the function evaluated at $x$ and the increment $\Delta x$:

\mathrm{d}y = f’(x) \cdot \Delta x \tag{1}

The differential of the independent variable $x$ is equal to the increment of the variable itself: $\mathrm{d}x = \Delta x.$ By substituting the value into the definition, we obtain:

\mathrm{d}y = f’(x) \cdot \mathrm{d}x \tag{2}

From the formula, it follows that the first derivative of a function is the ratio between the differential of the function and that of the independent variable:

f’(x) = \frac{\mathrm{d}y}{\mathrm{d}x} \tag{3}

From a geometric point of view, consider the triangle ABC. By the properties of trigonometry and of right triangles, the side $\overline{BC}$ can be rewritten as:

\overline{BC} = \overline{AB} \cdot \tan(\alpha) \tag{4}

where $\overline{AB} = \Delta x$ and $\tan(\alpha) = f’(x)$. The equality $(4)$ can therefore be rewritten as:

\begin{align} \overline{BC} &= \overline{AB} \cdot \tan(\alpha) \tag{5} \\[0.5em] &= \Delta x \cdot f’(x) \\[0.5em] &= \mathrm{d}y \end{align}

In other words, the differential $dy$ is the change in the ordinate of the tangent line to the curve when moving from point A with abscissa $x$ to point B with abscissa $x + \Delta x$.