Q. 34 - Calculus

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. 

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

Step-by-Step Solution

Answer

The required answer is $\frac{6x^{{\displaystyle \frac{5}{2}}}{(x^2-1)}^2-{(x^2-1)}^3\left(\frac{3}{2}x\frac{1}{2}\right)}{x^3}$

Step 1. Given Information

The given function is  $f\left(x\right)={\left(x^{-\frac{1}{2}}(x^2-1)\right)}^3$

Step 2. Calculation

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

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left({\displaystyle \frac{d}{dx}}{\left(x^2-1\right)}^3\right)x^{{\displaystyle \frac{3}{2}}}-{\left(x^2-1\right)}^3\left({\displaystyle \frac{d}{dx}}{x}^{{\displaystyle \frac{3}{2}}}\right)}{{\left(x^{{\displaystyle \frac{3}{2}}}\right)}^2} \\ =\frac{\left({\displaystyle 3{\left(x^2-1\right)}^{3-1}\frac{d}{dx}(x^2-1)}\right)x^{{\displaystyle \frac{3}{2}}}-{\left(x^2-1\right)}^3\left({\displaystyle \frac{3}{2}{x}^{\frac{3}{2}-1}}\right)}{\left(x^3\right)} \\ =\frac{\left({\displaystyle 3{\left(x^2-1\right)}^2\left(2x\right)}\right)x^{{\displaystyle \frac{3}{2}}}-{\left(x^2-1\right)}^3\left({\displaystyle \frac{3}{2}{x}^{\frac{1}{2}}}\right)}{\left(x^3\right)} \\ =\frac{6x^{\frac{5}{2}}{\left(x^2-1\right)}^2-{\left(x^2-1\right)}^3\left({\displaystyle \frac{3}{2}{x}^{\frac{1}{2}}}\right)}{\left(x^3\right)} \end{gathered}$$