Question

In example (2c) we calculated a derivative two different ways use algebra to simplify the first answer and show that it is equal to the second

Short answer

We simplify the first part andd prove that it is equal to derivative calculated by using second method

Step-by-step solution

Given information

We are given the derivative as

$h\text{'}\left(x\right)=\frac{(7x^6+{\displaystyle \frac{1}{2}}{x}^{-{\displaystyle \frac{1}{2}}})\left(x^3\right)-(x^7+\sqrt{x})\left(3x^2\right)}{x^6}$

Simplify

We have,

$$\begin{gathered} h\text{'}\left(x\right)=\frac{(7x^6+{\displaystyle \frac{1}{2}}{x}^{-{\displaystyle \frac{1}{2}}})\left(x^3\right)-(x^7+\sqrt{x})\left(3x^2\right)}{x^6} \\ h\text{'}\left(x\right)=\frac{7x^9+{\displaystyle \frac{1}{2}}{x}^{{\displaystyle \frac{5}{2}}}-3x^9-3x^{{\displaystyle \frac{5}{2}}}}{x^6} \\ h\text{'}\left(x\right)=\frac{4x^9-{\displaystyle \frac{5}{2}}{x}^{{\displaystyle \frac{5}{2}}}}{x^6} \\ h\text{'}\left(x\right)=4x^3-\frac{5}{2}{x}^{-\frac{7}{2}} \end{gathered}$$