Q. 1 - Calculus

Question

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

Step-by-Step Solution

Answer

$1. f\text{'}\text{'}\left(x\right) = \frac{2\left(9x^3-27x^2+18x-2\right)}{{\left(3x^2-4x+2\right)}^3}.$

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

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

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

Step 1. Given Information.

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

Step 2. Calculating the first derivative part(a).

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

Step 3. Calculating the second derivative part(a).

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

Step 4. Calculating the first derivative part(b).

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

Step 5. Calculating the second derivative part(b).

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

Step 6. Calculating the first derivative part(c).

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

Step 7. Calculating the second derivative part(c).

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

Step 8. Calculating the first derivative part(d).

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

Step 9. Calculating the second derivative part(d).

$$\begin{gathered} f\text{'}\left(x\right) = \frac{2x{e}^{3x}}{x^2+1} + 3f\left(x\right) \\ Differentiating once again, \\ as we know the derivative of f\left(x\right) we will find only the first term, \\ f\text{'}\text{'}\left(x\right) = 3f\text{'}\left(x\right) + \frac{\left(x^2+1\right){\displaystyle \frac{d}{dx}}\left(2x{e}^{3x}\right)-2x{e}^{3x}{\displaystyle \frac{d}{dx}}\left(x^2+1\right)}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right) = 3f\text{'}\left(x\right) + \frac{\left(x^2+1\right)\left(2e^{3x}+2x3e^{3x}\right)-2x{e}^{3x}\left(2x\right)}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right) = 3f\text{'}\left(x\right) + \frac{2e^{3x}\left(x^2+1\right)\left(1+3x\right)-4x^2{e}^{3x}}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right) = 3f\text{'}\left(x\right) + \frac{2e^{3x}\left(\left(3x^3+x^2+3x+1\right)-2x^2\right)}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right) = 3\left(\frac{2x{e}^{3x}}{x^2+1} + 3e^{3x}\ln \left(x^2+1\right)\right) + \frac{2e^{3x}\left(3x^3-x^2+3x+1\right)}{{\left(x^2+1\right)}^2}. \end{gathered}$$