hw_2.1_basic derivative rules\ndue sunday by 11:59pm points 100.05 submitting an exte\nhw_2.1_basic…

hw_2.1_basic derivative rules\ndue sunday by 11:59pm points 100.05 submitting an exte\nhw_2.1_basic derivative rules\n16. submit answer practice similar\nattempt 1: 10 attempts remaining.\nthe cost function of a certain commodity is (c(x)=82 + 0.16x-0.0006x^{2}+0.000003x^{3}). find (c(100)).\n(c(100)=)

hw_2.1_basic derivative rules\ndue sunday by 11:59pm points 100.05 submitting an exte\nhw_2.1_basic derivative rules\n16. submit answer practice similar\nattempt 1: 10 attempts remaining.\nthe cost function of a certain commodity is (c(x)=82 + 0.16x-0.0006x^{2}+0.000003x^{3}). find (c(100)).\n(c(100)=)

Answer

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:

$0.13$