Q. 52 - Calculus

Question

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

$\frac{d}{dx}{\overline{)\left(\frac{x^2+1}{\left(x^2+4\right)\left(3x-2\right)}\right)}}_{x=-1}$

Step-by-Step Solution

Answer

$\frac{d}{dx}{\overline{)\left(\frac{x^2+1}{\left(x^2+4\right)\left(3x-2\right)}\right)}}_{x=-1}$is $\frac{-24}{625}$

Step 1. Given Information

We need to find $\frac{d}{dx}{\overline{)\left(\frac{x^2+1}{\left(x^2+4\right)\left(3x-2\right)}\right)}}_{x=-1}$

Let us write $y=\frac{x^2+1}{\left(x^2+4\right)\left(3x-2\right)}$ 

Step 2. Finding First Derivative

$$\begin{gathered} \frac{dy}{dx}=\frac{\left(x^2+4\right)\left(3x-2\right){\displaystyle \frac{d}{dx}}\left(x^2+1\right)-\left(x^2+1\right){\displaystyle \frac{d}{dx}}\left(x^2+4\right)\left(3x-2\right)}{{\left(\left(x^2+4\right)\left(3x-2\right)\right)}^2} \\ \frac{dy}{dx}=\frac{\left(x^2+4\right)\left(3x-2\right)\left(2x\right)-\left(x^2+1\right)\left[2x\left(3x-2\right)+\left(x^2+4\right)3\right]}{{\left(x^2+4\right)}^2{\left(3x-2\right)}^2} \\ \frac{dy}{dx}=\frac{\left(x^2+4\right)\left(6x^2-4x\right)-\left(x^2+1\right)\left[6x^2-4x+3x^2+12\right]}{{\left(x^2+4\right)}^2{\left(3x-2\right)}^2} \\ \frac{dy}{dx}=\frac{\left(x^2+4\right)\left(6x^2-4x\right)-\left(x^2+1\right)\left(9x^2-4x+12\right)}{{\left(x^2+4\right)}^2{\left(3x-2\right)}^2}=\frac{-3x^4+3x^2+12x-12}{{\left(x^2+4\right)}^2{\left(3x-2\right)}^2} \end{gathered}$$

Step 3. Finding derivative at x = - 1

$$\begin{gathered} \frac{dy}{dx}=\frac{-3{\left(-1\right)}^4+3{\left(-1\right)}^2+12\left(-1\right)-12}{{\left(1+4\right)}^2{\left(-3-2\right)}^2} \\ \frac{dy}{dx}=\frac{-3+3-12-12}{{\left(1+4\right)}^2{\left(-3-2\right)}^2}=\frac{-24}{625} \end{gathered}$$