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.

$f\left(x\right)=\frac{x^2-\sqrt[3]{x}+5x^9}{\sqrt{x^{-1}}}$

Short answer

The required answer is$\frac{\left(2x-{\displaystyle \frac{3}{2}}{x}^{{\displaystyle \frac{1}{2}}}+45x^8\right)\sqrt{x^{-1}}+\left(x^2-\sqrt[3]{x}+5x^9\right)\left({\displaystyle \frac{1}{2}}{x}^{-{\displaystyle \frac{3}{2}}}\right)}{x^{-1}}$

Step-by-step solution

Step 1. Given Information   

The given function is$f\left(x\right)=\frac{x^2-\sqrt[3]{x}+5x^9}{\sqrt{x^{-1}}}$

Step 2. Calculation    

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

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left({\displaystyle \frac{d}{dx}}\left(x^2-\sqrt[3]{x}+5x^9\right)\right)\sqrt{x^{-1}}-\left(x^2-\sqrt[3]{x}+5x^9\right)\left({\displaystyle \frac{d}{dx}}\sqrt{x^{-1}}\right)}{{\left(\sqrt{x^{-1}}\right)}^2} \\ =\frac{\left({\displaystyle 2x-\frac{3}{2}{x}^{\frac{3}{2}-1}+5\left(9x^8\right)}\right)\sqrt{x^{-1}}-\left(x^2-\sqrt[3]{x}+5x^9\right)\left({\displaystyle -\frac{1}{2}{x}^{-\frac{1}{2}-1}}\right)}{x^{-1}} \\ =\frac{\left({\displaystyle 2x-\frac{3}{2}{x}^{\frac{1}{2}}+45x^8}\right)\sqrt{x^{-1}}-\left(x^2-\sqrt[3]{x}+5x^9\right)\left({\displaystyle -\frac{1}{2}{x}^{-\frac{3}{2}}}\right)}{x^{-1}} \\ =\frac{\left({\displaystyle 2x-\frac{3}{2}{x}^{\frac{1}{2}}+45x^8}\right)\sqrt{x^{-1}}+\left(x^2-\sqrt[3]{x}+5x^9\right)\left({\displaystyle \frac{1}{2}{x}^{-\frac{3}{2}}}\right)}{x^{-1}} \end{gathered}$$