Q.32 - 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.

$f\left(x\right)=\sqrt{x}(x^{-1}+1)$

Step-by-Step Solution

Answer

The derivative of the given function is $-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Step1. Given information

Here $f\left(x\right)=\sqrt{x}(x^{-1}+1)$

We have  to differentiate  with the product rule and

then without rule, after that,  use algebra to show that  the answers are the same.

Step 2. Differentiate using the product rule

When we apply the product rule on the given function , then $$\begin{gathered} \mathrm{The} \mathrm{Product} \mathrm{Rule}: ( {\mathrm{f}}^{\text{'}}\mathrm{g}) \left(\mathrm{x}\right) = \mathrm{f}\text{'} \left(\mathrm{x}\right) \mathrm{g}\left(\mathrm{x}\right) + \mathrm{f}\left(\mathrm{x}\right) \mathrm{g} \text{'}\left(\mathrm{x}\right) \\ \mathrm{here} \mathrm{take} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x} } , \mathrm{we} \mathrm{can} \mathrm{rewrite} \sqrt{\mathrm{x}}={\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{g}\left(\mathrm{x}\right)=\left({\mathrm{x}}^{-1}+1\right) , \\ \mathrm{Then} \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} , \mathrm{since} \mathrm{power} \mathrm{rule} : \mathrm{for} \mathrm{n} \frac{\mathrm{d}\left({\mathrm{x}}^{\mathrm{n}}\right)}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1} \\ \mathrm{g}\text{'}\left(\mathrm{x}\right)= -1\times {\mathrm{x}}^{-1-1} , \mathrm{constant} \mathrm{rule} \mathrm{and} \mathrm{power} \mathrm{rule} \mathrm{applied} \\ =-{\mathrm{x}}^{-2} \\ \mathrm{now} \mathrm{apply} \mathrm{the} \mathrm{product} \mathrm{rule} \mathrm{we} \mathrm{get} \\ \frac{d\left(\sqrt{\mathrm{x}}\right(\left({\mathrm{x}}^{-1}+1\right)}{d\mathrm{x}}=\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \left({\mathrm{x}}^{-1}+1\right)+\sqrt{\mathrm{x} }\times -{\mathrm{x}}^{-2} \\ =\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}({\mathrm{x}}^{-1}+1)-{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\mathrm{x}}^{-2} \end{gathered}$$

Step 3. Differentiate with out using product rule

Here function is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}({\mathrm{x}}^{-1}+1) \\ \mathrm{We} \mathrm{can} \mathrm{write} \sqrt{\mathrm{x}} \mathrm{as} {\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \mathrm{and} \mathrm{apply} \mathrm{distribution} \mathrm{law} \mathrm{on} \mathrm{the} \mathrm{given} \mathrm{function} \\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\mathrm{x}}^{-1}+{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} , \mathrm{we} \mathrm{know} \mathrm{that} {\mathrm{x}}^{\mathrm{m}}{\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}+\mathrm{n}} , \mathrm{then} \\ ={\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{now} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \frac{\mathrm{d}\left(\mathrm{f}\right(\mathrm{x}\left)\right)}{\mathrm{dx}}=\frac{\mathrm{d}({\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})}{\mathrm{dx}} \\ =\frac{\mathrm{d}\left({\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left({\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\mathrm{dx}} \mathrm{sum} \mathrm{rule} \mathrm{applied} \mathrm{here} \\ =-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1} \mathrm{since} \mathrm{power} \mathrm{rule} \frac{\mathrm{d}\left({\mathrm{x}}^{\mathrm{n}}\right)}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1} \\ =-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Step 4. Checking both the answers are the same

By using product rule derivatives of the  function is 

$$\begin{gathered} \frac{d\left(\sqrt{\mathrm{x}}\right(\left({\mathrm{x}}^{-1}+1\right)}{d\mathrm{x}}=\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}({\mathrm{x}}^{-1}+1)-{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\mathrm{x}}^{-2} , \mathrm{first} \mathrm{using} \mathrm{the} \mathrm{distributive} \mathrm{property} \mathrm{we} \mathrm{can} \mathrm{we} \mathrm{arrange} \mathrm{and} \mathrm{then} \mathrm{solve} , \mathrm{then} \mathrm{we} \mathrm{get} \\ = \frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\mathrm{x}}^{-1}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\mathrm{x}}^{-2} \\ =\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2 $}\right.} \mathrm{since} {\mathrm{x}}^{\mathrm{m}}{\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}+\mathrm{n}} \\ =-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

without applying the product rule derivatives of the function is

 $\frac{d\left(\sqrt{\mathrm{x}}\right(\left({\mathrm{x}}^{-1}+1\right)}{d\mathrm{x}}=-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Here using the algebra both answers are same.