Question

hw_2.1_basic derivative rules
due sunday by 11:59pm points 100.05 submitting an exte
hw_2.1_basic derivative rules

1. submit answer practice similar

attempt 1: 10 attempts remaining.
the cost function of a certain commodity is (c(x)=82 + 0.16x-0.0006x^{2}+0.000003x^{3}). find (c(100)).
(c(100)=)

Explanation:

Step1: Apply power - rule for derivatives

The power - rule states that if $y = ax^n$, then $y'=nax^{n - 1}$, and the derivative of a constant $c$ is $0$. Given $C(x)=82 + 0.16x-0.0006x^{2}+0.000003x^{3}$.
The derivative $C'(x)$ is:
$C'(x)=\frac{d}{dx}(82)+\frac{d}{dx}(0.16x)-\frac{d}{dx}(0.0006x^{2})+\frac{d}{dx}(0.000003x^{3})$
Since $\frac{d}{dx}(82) = 0$, $\frac{d}{dx}(0.16x)=0.16$, $\frac{d}{dx}(0.0006x^{2})=2\times0.0006x = 0.0012x$, and $\frac{d}{dx}(0.000003x^{3})=3\times0.000003x^{2}=0.000009x^{2}$.
So $C'(x)=0 + 0.16-0.0012x + 0.000009x^{2}$.

Step2: Evaluate $C'(x)$ at $x = 100$

Substitute $x = 100$ into $C'(x)$:
$C'(100)=0.16-0.0012\times100+0.000009\times(100)^{2}$
$C'(100)=0.16 - 0.12+0.09$
$C'(100)=0.13$.

Answer:

$0.13$