Q.33 - 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.

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

Step-by-Step Solution

Answer

The derivative of the given function is $\frac{3}{2}{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{5}{2}{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Step1. Given information

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

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

 Quotient rule states that  

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

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

Step 3. Differentiate with out using quotient rule

Based on question  function

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-{\mathrm{x}}^3}{\sqrt{\mathrm{x}}} , \mathrm{we} \mathrm{can} \mathrm{write} \sqrt{\mathrm{x}}={\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} , \mathrm{then} \frac{1}{\sqrt{\mathrm{x}}}={\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} , \mathrm{then} \mathrm{the} \mathrm{function} \mathrm{becomes} \\ =({\mathrm{x}}^2-{\mathrm{x}}^3){\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \mathrm{since} \mathrm{parenthesis} \mathrm{opened} \mathrm{by} \mathrm{distributive} \mathrm{law} \\ ={\mathrm{x}}^2{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^3{\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ ={\mathrm{x}}^{2-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{3-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \mathrm{since} {\mathrm{x}}^{\mathrm{m}}\times {\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}+\mathrm{n}} \\ \mathrm{f}\left(\mathrm{x}\right) ={\mathrm{x}}^{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{now} \mathrm{we} \mathrm{can} \mathrm{differentiate} \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{power} \mathrm{rule} \mathrm{of} \mathrm{differentiation} \mathrm{we} \mathrm{get} \\ \frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right) \\ =\frac{3}{2}{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1}-\frac{5}{2}{\mathrm{x}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1} \mathrm{since} \mathrm{the} \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}}\right)={\mathrm{nx}}^{\mathrm{n}-1} \\ =\frac{3}{2}{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{5}{2}{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$ 

Step 4. Checking both the answers are the same

 By using quotient rule derivatives of the function is 

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{x}}^2-{\mathrm{x}}^3}{\sqrt{\mathrm{x}}}\right)=\frac{(2{\mathrm{x}}^1-3{\mathrm{x}}^2){\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\displaystyle \frac{1}{2}}({\mathrm{x}}^1-{\mathrm{x}}^3){\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\mathrm{x}} \\ =\left(\right(2\mathrm{x}-3{\mathrm{x}}^2){\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}(\mathrm{x}-{\mathrm{x}}^3){\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} ) x^{-1} we know that \frac{1}{x}=x^{-1 } and to simplify we have to open the parenthesis ,u\sin g distributive law \\ =(2\mathrm{x}\times {\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-3{\mathrm{x}}^2\times {\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\mathrm{x}\times {\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^3\times {\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}){\mathrm{x}}^{-1} \\ =2\mathrm{x}\times {\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\mathrm{x}}^{-1}-3{\mathrm{x}}^2\times {\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\mathrm{x}}^{-1}-\frac{1}{2}\mathrm{x}\times {\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^3\times {\mathrm{x}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\mathrm{x}}^{-1} \sin ce x^m\times x^n=x^{m+n} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{x}}^7-{\mathrm{x}}^3}{\sqrt{\mathrm{x}}}\right) =2{\mathrm{x}}^{1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1}-3{\mathrm{x}}^{2+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1}-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$1-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{3-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1} \\ =2{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-3{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{1}{2}{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\frac{3}{2}{\mathrm{x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{5}{2}{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

   derivatives of the function without the applying the quotient rule is 

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{x}}^7-{\mathrm{x}}^3}{\sqrt{\mathrm{x}}}\right)=\frac{3}{2}{x}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{5}{2}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

  here by using the algebra both answers are same.