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.

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

Short answer

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

Step-by-step solution

Step 1. Given Information  

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

Step 2. Calculation     

Differentiate both the sides with respect tox, we get,

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