Chapter 3.2, Problem 2E

Find the derivative o f the function

$F(x)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$

in two ways: by using the Quotient Rule and by simplifying first. Show that your answers are equivalent. Which method do you prefer?

To determine

To find: The derivative of the function $F\left(x\right)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$.

Answer to Problem 2E

The derivative of the function $F\left(x\right)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$ is. $2x-5-\frac{3}{2}{x}^{\frac{-5}{2}}$.

Explanation of Solution

Given:

The function $F\left(x\right)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$.

Derivative rule:

(1) Quotient Rule: If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[\frac{f_1\left(x\right)}{f_2\left(x\right)}\right]=\frac{f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]-f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]}{{\left[f_2\left(x\right)\right]}^2}$

(2) Power Rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(3). Sum rule: $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$

(4) Constant multiple rule: $\frac{d}{dx}\left(cf\right)=c\frac{d}{dx}\left(f\right)$

(5) Difference rule: $\frac{d}{dx}\left(f-g\right)=\frac{d}{dx}\left(f\right)-\frac{d}{dx}\left(g\right)$

Calculation:

Method 1:

Obtain the derivative of $F\left(x\right)$ by using Quotient Rule.

The derivative of the function $F\left(x\right)$ is $F^{\prime }\left(x\right)$, which is obtained as follows.

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{d}{dx}\left(F\left(x\right)\right)\\ =\frac{d}{dx}\left(\frac{x^4-5x^3+\sqrt{x}}{x^2}\right)\end{array}$

Substitute $x^4-5x^3+\sqrt{x}$ for $f_1\left(x\right)$ and $x^2$ for $f_2\left(x\right)$ in the quotient rule (1),

$F^{\prime }\left(x\right)=\frac{\left(x^2\right)\frac{d}{dx}\left(x^4-5x^3+\sqrt{x}\right)-\left(x^4-5x^3+\sqrt{x}\right)\frac{d}{dx}\left(x^2\right)}{{\left(x^2\right)}^2}$

Apply the derivative rules (3), (4), and (5),

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{\left(x^2\right)\left[\frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(5x^3\right)+\frac{d}{dx}\left(x^{\frac{1}{2}}\right)\right]-\left[\left(x^4-5x^3+\sqrt{x}\right)\frac{d}{dx}\left(x^2\right)\right]}{{\left(x^2\right)}^2}\\ =\frac{x^2\left[\frac{d}{dx}\left(x^4\right)-5\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(x^{\frac{1}{2}}\right)\right]-\left[\left(x^4-5x^3+x^{\frac{1}{2}}\right)\frac{d}{dx}\left(x^2\right)\right]}{x^4}\end{array}$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{x^2\left[4x^3-5\left(3x^{3-1}\right)+\left(\frac{1}{2}{x}^{\frac{1}{2}-1}\right)\right]-\left(x^4-5x^3+x^{\frac{1}{2}}\right)2x}{x^4}\\ =\frac{x^2\left[4x^3-15x^2+\frac{1}{2}{x}^{\frac{-1}{2}}\right]-\left(x^4-5x^3+x^{\frac{1}{2}}\right)2x}{x^4}\\ =\frac{4x^5-15x^4+\frac{1}{2}{x}^{2-\frac{1}{2}}-2x^5+10x^4-2x^{1+\frac{1}{2}}}{x^4}\\ =\frac{4x^5-15x^4+\frac{1}{2}{x}^{\frac{3}{2}}-2x^5+10x^4-2x^{\frac{3}{2}}}{x^4}\end{array}$

Simplify the numerator and obtain the differentiation of $F\left(x\right)$.

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{2x^5-5x^4-\frac{3}{2}{x}^{\frac{3}{2}}}{x^4}\\ =2x-5-\frac{3}{2}{x}^{\frac{3}{2}-4}\\ =2x-5-\frac{3}{2}{x}^{\frac{-5}{2}}\end{array}$

Therefore, the differentiation of the function $F\left(x\right)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$ is $2x-5-\frac{3}{2}{x}^{\frac{-5}{2}}$.

Method 2:

Obtain the derivative of $F\left(x\right)$ by simplifying first.

Simplify the function $F\left(x\right)$,

$\begin{array}{c}F\left(x\right)=\frac{x^4-5x^3+x^{\frac{1}{2}}}{x^2}\\ =\frac{x^4}{x^2}-5\frac{x^3}{x^2}+\frac{x^{\frac{1}{2}}}{x^2}\\ =x^2-5x+x^{\frac{1}{2}-2}\\ =x^2-5x+x^{\frac{-3}{2}}\end{array}$

The derivative of the function $F\left(x\right)$ is $F^{\prime }\left(x\right)$, which is obtained as follows.

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{d}{dx}\left(F\left(x\right)\right)\\ =\frac{d}{dx}\left(x^2-5x+x^{\frac{-3}{2}}\right)\end{array}$

Apply the derivative rules (3), (4) and (5),

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(5x\right)+\frac{d}{dx}\left(x^{\frac{-3}{2}}\right)\\ =\frac{d}{dx}\left(x^2\right)-5\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^{\frac{-3}{2}}\right)\end{array}$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{F}^{\prime }\left(x\right)=2x-5\left(1\right)-\frac{3}{2}{x}^{\frac{-3}{2}-1}\\ =2x-5-\frac{3}{2}{x}^{\frac{-5}{2}}\end{array}$

Thus, the differentiation of the function $F\left(x\right)=\frac{x^4-5x^3+\sqrt{x}}{x^2}$ is $2x-5-\frac{3}{2}{x}^{\frac{-5}{2}}$.

Therefore, it can be concluded that the derivative of $F\left(x\right)$ by both the methods 1 and 2 are same and method 2 seems to be the better one as it has lesser calculations.