Q.31 - Calculus

Question

Differentiate each of the functions in Exercises 29–34 in two

different ways: first with the product and/or quotient rules and

then without these rules. Then use algebra to show that your

answers are the same.

$f\left(x\right)=x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(2-5x^3)$

Step-by-Step Solution

Answer

The derivatives of the given function is $7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{65}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Step1. Given information

Here $f\left(x\right)=x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(2-5x^3)$

We need to differentiate first with the product rule and

then without these rules, after that, we have to use algebra to show that  the answers are the same.

Step 2. Differentiate using the product rule

$\mathrm{The} \mathrm{Product} \mathrm{Rule}: ( {\mathrm{f}}^{\text{'}}\mathrm{g}) \left(\mathrm{x}\right) = \mathrm{f}\text{'} \left(\mathrm{x}\right) \mathrm{g}\left(\mathrm{x}\right) + \mathrm{f}\left(\mathrm{x}\right) \mathrm{g} \text{'}\left(\mathrm{x}\right)$

 when we differentiate using the product rule we get 

$$\begin{gathered} \mathrm{Product} \mathrm{Rule}: ( {\mathrm{f}}^{\text{'}}\mathrm{g}) \left(\mathrm{x}\right) = \mathrm{f}\text{'} \left(\mathrm{x}\right) \mathrm{g}\left(\mathrm{x}\right) + \mathrm{f}\left(\mathrm{x}\right) \mathrm{g} \text{'}\left(\mathrm{x}\right) \\ let f\left(\mathrm{x}\right)={\mathrm{x}}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \mathrm{and} \mathrm{g}\left(\mathrm{x}\right)=(2-5x^3) \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}\left({\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}{\mathrm{dx}} \\ =\frac{7}{2}{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} , \mathrm{since} \mathrm{based} \mathrm{on} \mathrm{power} \mathrm{rule} \frac{\mathrm{d}\left({\mathrm{x}}^{\mathrm{n}}\right)}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}} \\ \mathrm{g}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}(2-5x^3)}{\mathrm{dx}} \\ =\frac{\mathrm{d}\left(2\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(5x^3\right)}{\mathrm{dx}} \\ =0-15x^2 \\ \mathrm{then} \mathrm{based} \mathrm{on} \mathrm{product} \mathrm{rule} \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{function} \frac{d\left({\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right(2-5x^3)}{d\mathrm{x}}= \frac{7}{2}{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(2-5x^3)+{\mathrm{x}}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(-15{\mathrm{x}}^2 ) \\ =7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{35}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-15{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Step 3. Differentiate with out using product rule

Here  first, we have to use the distributive property, so  multiply by  $x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$   and open parenthesis  we get

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(2-5x^3) \\ \mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-5{\mathrm{x}}^{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{now} \mathrm{we} \mathrm{can} \mathrm{find} \mathrm{the} \mathrm{derivative} \mathrm{using} \mathrm{power} \mathrm{rule} ,\mathrm{for} \mathrm{any} \mathrm{non} \mathrm{rational} \mathrm{integer} \mathrm{n}, \frac{d{\mathrm{x}}^{\mathrm{n}}}{d\mathrm{x}}={\mathrm{nx}}^{\mathrm{n}-1} \\ \mathrm{then} \frac{d}{d\mathrm{x}}(2{\mathrm{x}}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-5{\mathrm{x}}^{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})=\frac{d\left(2{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}{d\mathrm{x}}+\frac{d(-5{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}})}{d\mathrm{x}} \\ =2\times \frac{7}{2}{\mathrm{x}}^{\frac{7}{2}-1}-5\times \frac{13}{2}{\mathrm{x}}^{\frac{13}{2}-1} \\ =7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{65}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{So} \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{is} 7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{65}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Step 4. Checking both the answers are the same

  without using the product rule  the derivative of  the given function is $7x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$$}\right.2}-5x^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

When we apply the product rule  the derivative of the given function is $7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{35}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-15{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ and when solve like terms then derivates becomes

$7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{35}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-15{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=7{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{65}{2}{\mathrm{x}}^{\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

So by using algebra both answers are same.