Q.30 - 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.

$f\left(x\right)=\frac{3x+1}{x^4}$

Step-by-Step Solution

Answer

 The derivatives for the given function is $-9x^{-4}-4x^{-5}$

Step1. Given information

Here given function is $f\left(x\right)=\frac{3x+1}{x^4}$

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

 when we apply the quotient rule we get ,

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{3\mathrm{x}+1}{{\mathrm{x}}^4}\right)=\frac{{\displaystyle \frac{d(3\mathrm{x}+1)}{d\mathrm{x}}}\times {\mathrm{x}}^4-(3\mathrm{x}+1)\times {\displaystyle \frac{d\left({\mathrm{x}}^4\right)}{d\mathrm{x}}}}{{\left({\mathrm{x}}^4\right)}^2} \\ =\frac{3\times {\mathrm{x}}^4-(3\mathrm{x}+1)\times 4{\mathrm{x}}^3}{{\mathrm{x}}^8} \mathrm{since} \mathrm{we} \mathrm{applied} \mathrm{identity} \mathrm{rule} \mathrm{and} \mathrm{constatnt} \mathrm{rule} \\ =\frac{3{\mathrm{x}}^4-12{\mathrm{x}}^4-4{\mathrm{x}}^3}{{\mathrm{x}}^8} \\ =\frac{-9{\mathrm{x}}^4-4{\mathrm{x}}^3}{{\mathrm{x}}^8} \end{gathered}$$

Step 3. Differentiate with out using quotient rule

Here the function is $f\left(x\right)=\frac{3x+1}{x^4}$ and we can rewrite the function as $f\left(x\right)=(3x+1)\times x^{-4}$. By applying distributive law  we can open the parenthesis  and solve then, the function becomes

$$\begin{gathered} f\left(x\right)=(3x+1)\times x^{-4} \\ =3x\times x^{-4}+x^{-4} \sin ce x^m\times x^n=x^{m-n} \\ =3x^{-3}+x^{-4} \end{gathered}$$

 then derivative of this function  $$\begin{gathered} \frac{\mathrm{d}(3{\mathrm{x}}^{-3}+{\mathrm{x}}^{-4})}{\mathrm{dx}}=\frac{\mathrm{d}\left(3{\mathrm{x}}^{-3}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left({\mathrm{x}}^{-4}\right)}{\mathrm{dx}} \\ =3\times -3\times {\mathrm{x}}^{-3-1}+(-4)\times {\mathrm{x}}^{-4-1} \mathrm{since} \mathrm{by} \mathrm{using} \mathrm{power} \mathrm{rule} \frac{\mathrm{d}\left({\mathrm{x}}^{\mathrm{n}}\right)}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1} \\ =-9{\mathrm{x}}^{-4}-4{\mathrm{x}}^{-5} \end{gathered}$$

Step 4. Checking both the answers are the same

$$\begin{gathered} \mathrm{Derivative} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{when} \mathrm{applying} \mathrm{quotient} \mathrm{rule} \mathrm{is} \frac{-9{\mathrm{x}}^4-12{\mathrm{x}}^3}{{\mathrm{x}}^8} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{3\mathrm{x}+1}{{\mathrm{x}}^4}\right)=\frac{-9{\mathrm{x}}^4-4{\mathrm{x}}^3}{{\mathrm{x}}^8} \mathrm{when} \mathrm{we} \mathrm{applying} \mathrm{exponent} \mathrm{rule}\raisebox{1ex}{$ {\mathrm{x}}^{\mathrm{m}}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{x}}^{\mathrm{n}}$}\right.={\mathrm{x}}^{\mathrm{m}-\mathrm{n}} \mathrm{on} \mathrm{the} \mathrm{n} \mathrm{derivative} \mathrm{function} , \mathrm{we} \mathrm{get} \\ =(-9{\mathrm{x}}^4-4{\mathrm{x}}^3)\times {\mathrm{x}}^{-8} \\ =-9{\mathrm{x}}^4\times {\mathrm{x}}^{-8}-4{\mathrm{x}}^3\times {\mathrm{x}}^{-8} \\ =-9{\mathrm{x}}^{4-8} -4 \times {\mathrm{x}}^{3-8} \mathrm{since} {\mathrm{x}}^{\mathrm{m}}\times {\mathrm{x}}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{m}+\mathrm{n}} \\ =-9{\mathrm{x}}^{-4}-4{\mathrm{x}}^{-5} \\ \mathrm{derivative} \mathrm{of} \mathrm{a} \mathrm{function} \mathrm{with} \mathrm{out} \mathrm{applying} \mathrm{quotient} \mathrm{is} -9{\mathrm{x}}^{-4}-4{\mathrm{x}}^{-5} \\ \mathrm{hence} \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{algebra} \mathrm{both} \mathrm{answers} \mathrm{are} \mathrm{same} . \end{gathered}$$