Question

Learn with an example Find the instantaneous rate of change of k(x) = x at x = 9.

Answer 1

Answer:

$k^{\prime}(9) = 1$.

Step-by-step explanation:

The expression $k(x) = x$ is equivalent to $k(x) = x^{1}$.

Apply the power rule of differentiation. For any constant $a$:

$\displaystyle \frac{d}{dx}[x^{a}] = a\, x^{a-1}$.

In $k(x) = x^{1}$, $a = 1$. Thus:

$\begin{aligned}k^{\prime}(x) &= \frac{d}{dx}[x^{1}] \\ &= 1\, x^{1 - 1} \\ &= x^{0} \\ &= 1 && \text{given that x \ne 0 }\end{aligned}$.

In other words, the instantaneous rate of change of $k(x)$ (with respect to $x$) is constantly $1$ at all $x \ne 0$.

Therefore, for $x = 9$, instantaneous rate of change of $k(x)$ (with respect to $x$) would be $1$.