Finding the second derivative: For each of the following functions $f$, calculate and simplify the second derivative $f\text{'}\text{'}$.

1. $f\left(x\right)=\frac{x-1}{3x^2-4x+2}.$

2. $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}.$

3. $f\left(x\right)=\frac{1-x}{e^x}.$

4.$f\left(x\right)=e^{3x}\ln \left(x^2+1\right).$

Given the function:

1.$f\left(x\right)=\frac{x-1}{3x^2-4x+2}.$

2.$f\left(x\right)=\frac{x}{\sqrt{x^2+1}}.$

3.$f\left(x\right)=\frac{1-x}{e^x}.$

4.$f\left(x\right)=e^{3x}\ln \left(x^2+1\right).$

$$\begin{gathered} f\left(x\right)=\frac{x-1}{3x^2-4x+2} \\ f\text{'}\left(x\right)=\frac{\left(3x^2-4x+2\right){\displaystyle \frac{d}{dx}}\left(x-1\right)-\left(x-1\right){\displaystyle \frac{d}{dx}}\left(3x^2-4x+2\right)}{{\left(3x^2-4x+2\right)}^2} \\ f\text{'}\left(x\right)=\frac{\left(3x^2-4x+2\right)-\left(x-1\right)\left(6x-4\right)}{{\left(3x^2-4x+2\right)}^2} \\ f\text{'}\left(x\right)=\frac{\left(3x^2-4x+2\right)-\left(6x^2-10x+4\right)}{{\left(3x^2-4x+2\right)}^2} \\ f\text{'}\left(x\right)=\frac{\left(-3x^2+6x-2\right)}{{\left(3x^2-4x+2\right)}^2} \end{gathered}$$

$$\begin{gathered} Nowdifferentiatingonceagain, \\ f\text{'}\text{'}\left(x\right)=\frac{\left({\left(3x^2-4x+2\right)}^2\right){\displaystyle \frac{d}{dx}}\left(-3x^2+6x-2\right)-\left(-3x^2+6x-2\right){\displaystyle \frac{d}{dx}}\left({\left(3x^2-4x+2\right)}^2\right)}{{\left(3x^2-4x+2\right)}^4} \\ f\text{'}\text{'}\left(x\right)=\frac{\left({\left(3x^2-4x+2\right)}^2\right)\left(-6x+6\right)-\left(-3x^2+6x-2\right)\left(2\left(3x^2-4x+2\right)\right)\left(6x-4\right)}{{\left(3x^2-4x+2\right)}^4} \\ f\text{'}\text{'}\left(x\right)=\frac{\left(\left(3x^2-4x+2\right)\right)\left(-6x+6\right)-\left(-3x^2+6x-2\right)\left(2\right)\left(6x-4\right)}{{\left(3x^2-4x+2\right)}^3} \\ f\text{'}\text{'}\left(x\right)=\frac{2\left(9x^3-27x^2+18x-2\right)}{{\left(3x^2-4x+2\right)}^3}. \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\frac{x}{\sqrt{x^2+1}} \\ f\text{'}\left(x\right)=\frac{\left(\sqrt{x^2+1}\right){\displaystyle \frac{d}{dx}}\left(x\right)-\left(x\right){\displaystyle \frac{d}{dx}}\left(\sqrt{x^2+1}\right)}{x^2+1} \\ f\text{'}\left(x\right)=\frac{\left(\sqrt{x^2+1}\right)-\left(x\right){\displaystyle \frac{2x}{2\sqrt{x^2+1}}}}{x^2+1} \\ f\text{'}\left(x\right)=\frac{\left(\sqrt{x^2+1}\right)-{\displaystyle \frac{x^2}{\sqrt{x^2+1}}}}{x^2+1} \\ f\text{'}\left(x\right)=\frac{\left(x^2+1\right)-{\displaystyle x^2}}{{\left(\sqrt{x^2+1}\right)}^3} \\ f\text{'}\left(x\right)=\frac{1}{{\left(\sqrt{x^2+1}\right)}^3}. \end{gathered}$$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{1}{{\left(\sqrt{x^2+1}\right)}^3}={\left(x^2+1\right)}^{-\frac{3}{2}} \\ Differentiatingonceagainweget, \\ f\text{'}\text{'}\left(x\right)=-\frac{3}{2}{\left(x^2+1\right)}^{-\frac{5}{2}}\left(2x\right)=\frac{-3x}{{\left(\sqrt{x^2+1}\right)}^5}. \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\frac{1-x}{e^x}. \\ f\text{'}\left(x\right)=\frac{e^x{\displaystyle \frac{d}{dx}}\left(1-x\right)-\left(1-x\right){\displaystyle \frac{d}{dx}}\left(e^x\right)}{e^{2x}} \\ f\text{'}\left(x\right)=\frac{e^x(-1)-(1-x)e^x}{e^{2x}}=\frac{(-1)-(1-x)}{e^x} \\ f\text{'}\left(x\right)=\frac{x-2}{e^x}. \end{gathered}$$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{x-2}{e^x} \\ Differentiatingonceagain, \\ f\text{'}\text{'}\left(x\right)=\frac{e^x{\displaystyle \frac{d}{dx}}(x-2)-(x-2){\displaystyle \frac{d}{dx}}\left(e^x\right)}{e^{2x}} \\ f\text{'}\text{'}\left(x\right)=\frac{e^x-(x-2)\left(e^x\right)}{e^{2x}} \\ f\text{'}\text{'}\left(x\right)=\frac{1-(x-2)}{e^x}=\frac{3-x}{e^x}. \end{gathered}$$

$$\begin{gathered} f\left(x\right)=e^{3x}\ln \left(x^2+1\right) \\ f\text{'}\left(x\right)=e^{3x}\frac{d}{dx}\left(\ln \left(x^2+1\right)\right)+\ln \left(x^2+1\right)\frac{d}{dx}\left(e^{3x}\right) \\ f\text{'}\left(x\right)=e^{3x}\frac{2x}{x^2+1}+\left(\ln \left(x^2+1\right)\right)\left(3e^{3x}\right) \\ f\text{'}\left(x\right)=\frac{2x{e}^{3x}}{x^2+1}+3f\left(x\right). \end{gathered}$$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{2x{e}^{3x}}{x^2+1}+3f\left(x\right) \\ Differentiatingonceagain, \\ asweknowthederivativeoff\left(x\right)wewillfindonlythefirstterm, \\ f\text{'}\text{'}\left(x\right)=3f\text{'}\left(x\right)+\frac{\left(x^2+1\right){\displaystyle \frac{d}{dx}}\left(2x{e}^{3x}\right)-2x{e}^{3x}{\displaystyle \frac{d}{dx}}\left(x^2+1\right)}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right)=3f\text{'}\left(x\right)+\frac{\left(x^2+1\right)\left(2e^{3x}+2x3e^{3x}\right)-2x{e}^{3x}\left(2x\right)}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right)=3f\text{'}\left(x\right)+\frac{2e^{3x}\left(x^2+1\right)\left(1+3x\right)-4x^2{e}^{3x}}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right)=3f\text{'}\left(x\right)+\frac{2e^{3x}\left(\left(3x^3+x^2+3x+1\right)-2x^2\right)}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right)=3\left(\frac{2x{e}^{3x}}{x^2+1}+3e^{3x}\ln \left(x^2+1\right)\right)+\frac{2e^{3x}\left(3x^3-x^2+3x+1\right)}{{\left(x^2+1\right)}^2}. \end{gathered}$$