Q.34 - 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.

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

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.}$

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  

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

$$\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 

$$\begin{gathered} f\left(x\right)=\frac{\sqrt{x}}{x^{-2}{x}^3} \\ we can write \sqrt{x}=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} , \\ by u\sin g laws of exponent x \end{gathered}$$

Step 4. Checking both the answers are the same

Without using the quotient rule derivative of the given function is 

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