Q. 48 - Calculus

Question

Calculate each of the derivatives or derivative values in Exercises 47–52.

$\frac{d^2}{d{x}^2}\left(\frac{{\displaystyle \frac{1}{x}-3x^2}}{x^5-{\displaystyle \frac{1}{\sqrt{x}}}}\right)$

Step-by-Step Solution

Answer

The second derivative of $\frac{d^2}{d{x}^2}\left(\frac{{\displaystyle \frac{1}{x}-3x^2}}{x^5-{\displaystyle \frac{1}{\sqrt{x}}}}\right)$ is $\frac{-393x^{{\displaystyle \raisebox{1ex}{$29$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-97x^{{\displaystyle \raisebox{1ex}{$23$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+45x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-3\sqrt{x}-144x^{20}+168x^{17}+492x^9-68x^6}{-16x^{{\displaystyle \raisebox{1ex}{$39$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-16x^{{\displaystyle \raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+4x^{25}+24x^{14}+4x^3}$

Step 1. Given Information

We need to ask second order derivative of $\frac{d^2}{d{x}^2}\left(\frac{{\displaystyle \frac{1}{x}-3x^2}}{x^5-{\displaystyle \frac{1}{\sqrt{x}}}}\right)$

Let us write $y=\left(\frac{{\displaystyle \frac{1}{x}-3x^2}}{x^5-{\displaystyle \frac{1}{\sqrt{x}}}}\right)$

Step 2. Finding First Order Derivative

$$\begin{gathered} \frac{d}{dx}\left(\frac{-3x^2+{\displaystyle \frac{1}{x}}}{x^5-{\displaystyle \frac{1}{\sqrt{x}}}}\right)=\left(\frac{{\displaystyle \frac{d}{dx}}\left(-3x^2+\frac{1}{x}\right)\left(x^5-\frac{1}{\sqrt{x}}\right)-\left(-3x^2+\frac{1}{x}\right){\displaystyle \frac{d}{dx}}\left(x^5-\frac{1}{\sqrt{x}}\right)}{{\left(x^5-\frac{1}{\sqrt{x}}\right)}^2}\right) \\ =\left(\frac{{\displaystyle \left(-6x-\frac{1}{x^2}\right)\left(x^5-\frac{1}{\sqrt{x}}\right)-\left(-3x^2+\frac{1}{x}\right)\left(5x^4+\frac{1}{2x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)}}{{\left(x^5-\frac{1}{\sqrt{x}}\right)}^2}\right)=\frac{-18x^{{\displaystyle \raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+12x^{{\displaystyle \raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-15x^3-1}{-2x^{{\displaystyle \raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-2x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+4x^7} \end{gathered}$$

Step 3. Finding second order derivative

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{\left(-2x^{\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+4x^7\right){\displaystyle \frac{d}{dx}}\left(-18x^{\raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+12x^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-15x^3-1\right)-\left(-18x^{\raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+12x^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-15x^3-1\right){\displaystyle \frac{d}{dx}}\left(-2x^{\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+4x^7\right)}{{\left(-2x^{\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+4x^7\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{\left(-2x^{\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+4x^7\right)\left(-153x^{15/2}+66x^{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-45x^2-0\right)-\left(-18x^{\raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+12x^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-15x^3-1\right)\left(-25x^{{\displaystyle \raisebox{1ex}{$23$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-3\sqrt{x}+28x^6\right)}{{\left(-2x^{\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+4x^7\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{-393x^{{\displaystyle \raisebox{1ex}{$29$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-97x^{{\displaystyle \raisebox{1ex}{$23$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+45x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-3\sqrt{x}-144x^{20}+168x^{17}+492x^9-68x^6}{-16x^{{\displaystyle \raisebox{1ex}{$39$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-16x^{{\displaystyle \raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+4x^{25}+24x^{14}+4x^3} \end{gathered}$$