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.

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

Short answer

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

Step-by-step solution

Step1. Given information  

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

We need to differentiate first with the quotient 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 quotient rule 

The quotient rule states that

$$\begin{gathered} \mathrm{If}\mathrm{f}\mathrm{and}\mathrm{g}\mathrm{are}\mathrm{functions}\mathrm{and}\mathrm{both}\mathrm{f}\mathrm{and}\mathrm{g}\mathrm{are}\mathrm{differentiable}, \\ \mathrm{then}\mathrm{quotient}\mathrm{rule}{\left(\frac{\mathrm{f}}{\mathrm{g}}\right)}^{\text{'}}\left(\mathrm{x}\right)=\frac{\mathrm{f}\left(\mathrm{x}\right)\text{'}\mathrm{g}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\text{'}}{(\mathrm{g}{\left(\mathrm{x}\right))}^2} \end{gathered}$$

localid="1649046997566" $$\begin{gathered} \mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}} \\ \mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{-2}{\mathrm{x}}^3 \\ \mathrm{When}\mathrm{we}\mathrm{apply}\mathrm{quotient}\mathrm{rule},\mathrm{we}\mathrm{get}\mathrm{derivative}\mathrm{function}\mathrm{as} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^{-2}{\mathrm{x}}^3}\right)=\frac{{\displaystyle \frac{d\left(\sqrt{\mathrm{x}}\right)}{d\mathrm{x}}\times \left({\mathrm{x}}^{-2}{\mathrm{x}}^3\right)-\sqrt{\mathrm{x}}\frac{d\left({\mathrm{x}}^{-2}{\mathrm{x}}^3\right)}{d\mathrm{x}}}}{{\left({\mathrm{x}}^{-2}{\mathrm{x}}^3\right)}^2} \\ =\frac{\left(\frac{\mathrm{d}}{\mathrm{dx}}\right({\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left)\right({\mathrm{x}}^{-2}{\mathrm{x}}^3)-\sqrt{\mathrm{x}}\times \frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^{{\displaystyle -2}}{\mathrm{x}}^3)}{{\left({\mathrm{x}}^{-2}{\mathrm{x}}^3\right)}^2}\mathrm{since}\sqrt{\mathrm{x}}={\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\frac{\left(\frac{\mathrm{d}}{\mathrm{dx}}\right({\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left)\right(\mathrm{x})-\sqrt{\mathrm{x}}\times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})}{{\left(\mathrm{x}\right)}^2}\mathrm{we}\mathrm{know}\mathrm{that}{\mathrm{x}}^{\mathrm{m}}\times {\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}+\mathrm{n}}\mathrm{so}{\mathrm{x}}^{-2}\times {\mathrm{x}}^3={\mathrm{x}}^1\mathrm{and}{\mathrm{x}}^1=\mathrm{x} \\ =\frac{{\displaystyle \frac{1}{2}\mathrm{x}\times {\mathrm{x}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\sqrt{\mathrm{x}}}}{{\left(\mathrm{x}\right)}^2}\mathrm{since}\mathrm{power}\mathrm{rule}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}}\right)={\mathrm{nx}}^{\mathrm{n}-1}\mathrm{so}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{1}{2}{\mathrm{x}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{and}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)=1 \end{gathered}$$

Step 3. Differentiate with out using quotient  rule 

given function is

$$\begin{gathered} f\left(x\right)=\frac{\sqrt{x}}{x^{-2}{x}^3} \\ wecanwrite\sqrt{x}=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}, \\ byu\sin glawsofexponentx \end{gathered}$$

Step 4. Checking  both the answers are the same

Without using the quotient rule derivative of the given function is

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^{-2}{\mathrm{x}}^3}\right)=-\frac{1}{2}{\mathrm{x}}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{then}\mathrm{by}\mathrm{applying}\mathrm{the}\mathrm{quotient}\mathrm{rule}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{unction}\mathrm{is} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^{-2}{\mathrm{x}}^3}\right)=\frac{{\displaystyle \frac{1}{2}\mathrm{x}\times {\mathrm{x}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\left(\mathrm{x}\right)}^2} \\ \mathrm{apply}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{exponents}\mathrm{and}\mathrm{solve}\mathrm{the}\mathrm{expression}\mathrm{becomes} \\ =\left(\frac{1}{2}{\mathrm{x}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1}-{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right){\mathrm{x}}^{-2}\mathrm{since}\frac{1}{{\mathrm{x}}^{\mathrm{m}}}={\mathrm{x}}^{-\mathrm{m}}\mathrm{and}{\mathrm{x}}^{\mathrm{m}}{\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}+\mathrm{n}} \\ =\frac{1}{2}{\mathrm{x}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1-2}-{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-2} \\ =\frac{1}{2}{\mathrm{x}}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =-\frac{1}{2}{\mathrm{x}}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{so}\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{algebra}\mathrm{both}\mathrm{answers}\mathrm{are}\mathrm{same}. \end{gathered}$$