Instantaneous rate of change
The instantaneous rate of change of a function at a specific point measures how the function's output is changing at that exact point with respect to its input. It is equal to the derivative of the function at that point.
For a function \(f\left(x\right)\), the instantaneous rate of change is given by the limit definition of the derivative:
\[\frac{dy}{dx} = \lim_{h\to0}\frac{f\left(x + h\right) - f\left(x\right)}{h}\]
Alternative notation
\[f^\prime\left(x\right) = \lim_{h\to0}\frac{f\left(x + h\right) - f\left(x\right)}{h}\]
\(\dfrac{dy}{dx}\) is read as “the derivative of \(y\) with respect to \(x\)”, or the instantaneous rate of change of the function \(y = f(x)\) as \(x\) changes. \(f^\prime\left(x\right)\) is read as “\(f\ prime\ of\ x\)”. This formula determines the slope of the tangent line at a given point \(x\) on the curve.
Worked Example
Find the rule for the general derivative of \(f\left(x\right) = x^2 + 2\).
Hence, find the derivative (slope) of the function when \(x = 2\).
Let \(f\left(x\right) = x^{2} + 2\). Then
\[\begin{aligned} f^\prime\left(x\right) & = \lim_{h\to0}\frac{f\left(x + h\right) - f\left(x\right)}{h} \\ & = \lim_{h\to0}\frac{\left(\left(x + h\right)^{2} + 2\right) - \left(x^{2} + 2\right)}{h} \\ & = \lim_{h\to0}\frac{\left(x^{2} + 2xh + h^{2} + 2\right) - \left(x^{2} + 2\right)}{h} \\ & = \lim_{h\to0}\frac{2xh + h^{2}}{h} \\ & = \lim_{h\to0}\frac{h\left(2x + h\right)}{h} \\ & = \lim_{h\to0}\left(2x + h\right) \\ & = 2x \end{aligned}\]
Substitute \(x=2\) into the derivative
\[f^\prime\left(2\right)=2 \times 2 = 4\]
Therefore, the general derivative is \(f^{\prime}(x)=2 x\), and the slope of the function at \(x=2\) is \(4\).
