Q. 29 - 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.  

$f\left(x\right)=\frac{{\displaystyle \frac{1}{x}}-3x^2}{x^5-{\displaystyle \frac{1}{\sqrt{x}}}}$

Step-by-Step Solution

Answer

The required answer is $\frac{(-x^2-6x)}{\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)}-\frac{\left({\displaystyle \frac{1}{x}}-3x^2\right)\left(5x^4+{\displaystyle \frac{1}{2}}{x}^{-{\displaystyle \frac{3}{2}}}\right)}{{\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)}^2}$

Step 1. Given Information

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

Step 2. Calculation

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

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left({\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{1}{x}}-3x^2\right)\right)\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)-\left(\frac{1}{x}-3x^2\right)\left({\displaystyle \frac{d}{dx}}\left(x^5-\frac{1}{\sqrt{x}}\right)\right)}{{\left(x^5-\frac{1}{\sqrt{x}}\right)}^2} \\ =\frac{\left({\displaystyle -x^{-1-1}-3\left(2x\right)}\right)\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)-\left(\frac{1}{x}-3x^2\right){\displaystyle \left(5x^4-\left(-\frac{1}{2}{x}^{-\frac{1}{2}-1}\right)\right)}}{{\left(x^5-\frac{1}{\sqrt{x}}\right)}^2} \\ =\frac{\left({\displaystyle -x^{-2}-6x}\right)\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)-\left(\frac{1}{x}-3x^2\right){\displaystyle \left(5x^4+\frac{1}{2}{x}^{-\frac{3}{2}}\right)}}{{\left(x^5-\frac{1}{\sqrt{x}}\right)}^2} \\ =\frac{(-x^2-6x)}{\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)}-\frac{\left({\displaystyle \frac{1}{x}-3x^2}\right)\left(5x^4+{\displaystyle \frac{1}{2}}{x}^{-{\displaystyle \frac{3}{2}}}\right)}{{\left(x^5-{\displaystyle \frac{1}{\sqrt{x}}}\right)}^2} \end{gathered}$$