A function y = f(x) is given. At some point the increment
A function $y = f(x)$ is given. At some point the increment of the independent variable is $\Delta x = 0.2$ and the corresponding differential is $df = 0.8$. Find the value of the derivative at that point.
Explanation of Method
The derivative as the ratio of two differentials.
Step 1: The definition
\[ df = f'(x)\,\Delta x. \]
For the independent variable the differential and the increment coincide, $dx = \Delta x$, so this can also be written $df = f'(x)\,dx$.
Step 2: Divide
Since $\Delta x = 0.2 \neq 0$: \[f'(x) = \frac{df}{\Delta x} = \frac{0.8}{0.2} = 4.\]
Step 3: Verification
\[df = 4\times0.2 = 0.8 \;\;\checkmark\]
\[\text{The function itself is irrelevant: any } f \text{ with slope } 4 \text{ there gives the same } df\]
\[\text{Problem 268 is the special case } f(x) = x^{2}, \text{ where } 2x = 4 \text{ locates the point } x = 2 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,f'(x) = 4\,}\]
Remark
Remark. This is the reason for the notation $\dfrac{dy}{dx}$. The derivative is genuinely the quotient of the differential of the function by the differential of the argument — not merely a symbol shaped like a quotient — and that is what makes formal manipulations such as $dy = y'\,dx$, or the chain rule written as $\dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx}$, legitimate rather than mnemonic.