Question

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.

$f\left(x\right)=\frac{(x+1)(3x-4)}{\sqrt{x^3-27}}$

Short answer

The required answer is$\frac{-7}{\left(\sqrt{x^3-27}\right)}-\frac{\left(x+1\right)\left(3x-4\right)\left({\displaystyle \frac{3x^2}{2\sqrt{x^3-27}}}\right)}{{\left(\sqrt{x^3-27}\right)}^2}$

Step-by-step solution

Step 1. Given Information   

The given function is$f\left(x\right)=\frac{(x+1)(3x-4)}{\sqrt{x^3-27}}$

Step 2. Calculation   

Differentiate both the sides with respect to x, we get,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left({\displaystyle \frac{d}{dx}}\left(\left(x+1\right)(3x-4\right)\right)\left(\sqrt{x^3-27}\right)-\left(\left(x+1\right)\left(3x-4\left({\displaystyle \frac{d}{dx}}\left(\sqrt{x^3-27}\right)\right)\right)\right)}{{\left(\sqrt{x^3-27}\right)}^2} \\ =\frac{\left({\displaystyle \frac{d}{dx}\left(\left(x+1\right)\right)}\right)(3x-4)-\left(x+1\right)\left({\displaystyle \frac{d}{dx}}\left(3x-4\right)\right)\left(\sqrt{x^3-27}\right)-\left(x+1\right)(3x-4)\left(\frac{1}{2}{\left(x^3-27\right)}^{-\frac{1}{2}}{\displaystyle \frac{d}{dx}}\left(x^3-27\right)\right)}{{\left(\sqrt{x^3-27}\right)}^2} \\ =\frac{\left((3x-4)-3\left(x+1\right)\right)\left(\sqrt{x^3-27}\right)-\left(x+1\right)(3x-4)\left({\displaystyle \frac{3x^2}{2\sqrt{x^3-27}}}\right)}{{\left(\sqrt{x^3-27}\right)}^2} \\ =\frac{\left(3x-4-3x-3\right)}{\left(\sqrt{x^3-27}\right)}-\frac{\left(x+1\right)(3x-4)\left({\displaystyle \frac{3x^2}{2\sqrt{x^3-27}}}\right)}{{\left(\sqrt{x^3-27}\right)}^2} \\ =\frac{-7}{\left(\sqrt{x^3-27}\right)}-\frac{\left(x+1\right)(3x-4)\left({\displaystyle \frac{3x^2}{2\sqrt{x^3-27}}}\right)}{{\left(\sqrt{x^3-27}\right)}^2} \end{gathered}$$