1 Answer

Tathagata · Answer 1 · 2022-05-27T03:39:50+0000

Answer:

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

Explanation:

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

Apply the power rule of differentiation. For any constant
:

$\displaystyle (d)/(dx)[x^(a)] = a\, x^(a-1)$ .

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

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

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

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

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