Q. 18 - Calculus

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 

Step-by-Step Solution

Answer

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

Step 1: Given information

We are given the derivative as

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

Step 2: Simplify

We have,

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