Differential of a Function

Definition

1.Consider a differentiable function $f(x)$ 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 $x + \Delta x \in [a,b]$, separated by the same increment that appears in the difference quotient.

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

$$\mathrm{d}y = f'(x) \cdot \Delta x$$

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

$$\mathrm{d}y = f'(x) \cdot \mathrm{d}x$$

4.From this 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:

$$f'(x) = \frac{\mathrm{d}y}{\mathrm{d}x}$$

5.

This relation justifies the Leibniz notation $\mathrm{d}y/\mathrm{d}x$ for the derivative, which in this context acquires the precise meaning of a quotient of differentials.

Geometric interpretation

6.The geometric meaning of the differential emerges by considering the tangent line to the graph of $f$ at the point $A(x, f(x))$. Let $B(x + \Delta x, f(x))$ be the point obtained by moving horizontally from $A$ by the increment $\Delta x$, and let $C$ be the point on the tangent line with abscissa $x + \Delta x$. The three points form a right triangle $ABC$, whose legs are the horizontal segment $\overline{AB}$ and the vertical segment $\overline{BC}$.

7.Denoting by $\alpha$ the angle that the tangent line forms with the horizontal direction, the properties of right triangle trigonometry give:

$$\overline{BC} = \overline{AB} \cdot \tan(\alpha) \tag{1}$$

8.In this triangle $\overline{AB} = \Delta x$, and the slope of the tangent line satisfies $\tan(\alpha) = f'(x)$. The equality $(1)$ can therefore be rewritten as:

$$ \begin{align} \overline{BC} &= \overline{AB} \cdot \tan(\alpha) \\[6pt] &= \Delta x \cdot f'(x) \\[6pt] &= \mathrm{d}y \end{align} $$

9.In other words, the differential $\mathrm{d}y$ is the change in the ordinate of the tangent line to the curve when moving from the point with abscissa $x$ to the point with abscissa $x + \Delta x$. The differential thus measures the increment predicted by the linear approximation of the function, while the true increment of the function is $\Delta y = f(x + \Delta x) - f(x)$. The relation between the two quantities can be written as:

$$\Delta y = \mathrm{d}y + o(\Delta x) \tag{2}$$

10.The remainder is a little-o term, so the difference $\Delta y - \mathrm{d}y$ tends to zero faster than $\Delta x$ as $\Delta x \to 0$. This is the precise sense in which the differential provides the best linear approximation of the function near the point.

The differential and the actual increment

11.The distinction between the differential and the true increment can be made concrete by examining the function $f(x) = x^2$. Its derivative is $f'(x) = 2x$, so the differential relative to the point $x$ and the increment $\Delta x$ is given by:

$$\mathrm{d}y = f'(x) \cdot \Delta x = 2x \cdot \Delta x$$

12.The actual increment of the function over the same interval is the difference between the values of $f$ at the two endpoints. Expanding the square, we obtain:

$$ \begin{align} \Delta y &= f(x + \Delta x) - f(x) \\[6pt] &= (x + \Delta x)^2 - x^2 \\[6pt] &= 2x \cdot \Delta x + (\Delta x)^2 \end{align} $$

13.Comparing the two expressions, the actual increment exceeds the differential by the quantity $(\Delta x)^2$, which is exactly the little-o remainder anticipated by formula $(2)$. As a numerical illustration, take $x = 3$ and $\Delta x = 0.1$. The differential gives $\mathrm{d}y = 2 \cdot 3 \cdot 0.1 = 0.6$, while the actual increment is $\Delta y = 2 \cdot 3 \cdot 0.1 + (0.1)^2 = 0.61$. The two values differ by $0.01$, that is, by the square of the increment, and this discrepancy shrinks rapidly as $\Delta x$ is reduced.

Application to approximate calculations

14.The fact that $\Delta y$ and $\mathrm{d}y$ differ by a negligible quantity for small increments leads to a practical method for estimating the value of a function near a point where it is easily computed. Starting from $\Delta y = f(x + \Delta x) - f(x)$ and replacing the actual increment with the differential, we obtain the approximation:

$$f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$$

15.The procedure is effective when the base point $x$ is chosen so that both $f(x)$ and $f'(x)$ are simple to evaluate, and the increment $\Delta x$ is small. As an example, consider the estimate of $\sqrt{4.05}$ obtained from the function $f(x) = \sqrt{x}$, whose derivative is:

$$f'(x) = \frac{1}{2\sqrt{x}}$$

16.Choosing the base point $x = 4$, for which the square root is known exactly, and the increment $\Delta x = 0.05$, the derivative at the base point equals:

$$f'(4) = \frac{1}{2 \cdot 2} = \frac{1}{4}$$

17.Substituting these values into the approximation, we find:

$$ \begin{align} \sqrt{4.05} &\approx f(4) + f'(4) \cdot 0.05 \\[6pt] &= 2 + \frac{1}{4} \cdot 0.05 \\[6pt] &= 2.0125 \end{align} $$

18.The value returned by a direct computation is $\sqrt{4.05} = 2.012461\ldots$, so the estimate produced by the differential is accurate to four decimal places. The small error reflects the little-o remainder discarded when the actual increment was replaced by the differential.