report final answers rounded to three decimal places unless the context dictates otherwise. include units…

Explanation:

Step1: Recall derivative rules

Use power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, product - rule $\frac{d}{dx}(uv)=u'v + uv'$, $\frac{d}{dx}(a)=0$ for constant $a$, $\frac{d}{dx}(a\cdot f(x))=a\cdot\frac{d}{dx}(f(x))$, $\frac{d}{dx}(b^x)=b^x\ln b$, $\frac{d}{dx}(\ln x)=\frac{1}{x}$, $\frac{d}{dx}(e^x)=e^x$.

Step2: Derive $y = ax + b$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(ax + b)=a$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(a)=0$.

Step3: Derive $y = ax^{2}+bx + c$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(ax^{2}+bx + c)=2ax + b$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(2ax + b)=2a$.

Step4: Derive $y = ax^{3}+bx^{2}+cx + d$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(ax^{3}+bx^{2}+cx + d)=3ax^{2}+2bx + c$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(3ax^{2}+2bx + c)=6ax+2b$.

Step5: Derive $y = a\cdot b^{x}$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(a\cdot b^{x})=a\cdot b^{x}\ln b$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(a\cdot b^{x}\ln b)=a\cdot b^{x}(\ln b)^2$.

Step6: Derive $y = a + b\ln x$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(a + b\ln x)=\frac{b}{x}$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{b}{x})=-\frac{b}{x^{2}}$.

Step7: Derive $y = e^{x}$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(e^{x})=e^{x}$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(e^{x})=e^{x}$.

Step8: Derive $y = k\sqrt{x}=k x^{\frac{1}{2}}$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(k x^{\frac{1}{2}})=\frac{k}{2}x^{-\frac{1}{2}}$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{k}{2}x^{-\frac{1}{2}})=-\frac{k}{4}x^{-\frac{3}{2}}$.

Step9: Derive $y=\sqrt[3]{x}=x^{\frac{1}{3}}$

First - derivative: $\frac{dy}{dx}=\frac{d}{dx}(x^{\frac{1}{3}})=\frac{1}{3}x^{-\frac{2}{3}}$. Second - derivative: $\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{1}{3}x^{-\frac{2}{3}})=-\frac{2}{9}x^{-\frac{5}{3}}$.

Step10: Derive $y = f(x)\cdot g(x)$

First - derivative (product - rule): $\frac{dy}{dx}=f'(x)g(x)+f(x)g'(x)$. Second - derivative: $\frac{d^2y}{dx^2}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$.

Answer:

For $y = ax + b$: First - derivative is $a$, Second - derivative is $0$. For $y = ax^{2}+bx + c$: First - derivative is $2ax + b$, Second - derivative is $2a$. For $y = ax^{3}+bx^{2}+cx + d$: First - derivative is $3ax^{2}+2bx + c$, Second - derivative is $6ax + 2b$. For $y = a\cdot b^{x}$: First - derivative is $a\cdot b^{x}\ln b$, Second - derivative is $a\cdot b^{x}(\ln b)^2$. For $y = a + b\ln x$: First - derivative is $\frac{b}{x}$, Second - derivative is $-\frac{b}{x^{2}}$. For $y = e^{x}$: First - derivative is $e^{x}$, Second - derivative is $e^{x}$. For $y = k\sqrt{x}$: First - derivative is $\frac{k}{2}x^{-\frac{1}{2}}$, Second - derivative is $-\frac{k}{4}x^{-\frac{3}{2}}$. For $y=\sqrt[3]{x}$: First - derivative is $\frac{1}{3}x^{-\frac{2}{3}}$, Second - derivative is $-\frac{2}{9}x^{-\frac{5}{3}}$. For $y = f(x)\cdot g(x)$: First - derivative is $f'(x)g(x)+f(x)g'(x)$, Second - derivative is $f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$.