Math · Calculus I · Worked example

Simplify a quotient before using the power rule

Differentiate $f (x) = (x^{3} + 6) / x^{2}$ , then find the slope of its graph at $x = 2$ .

f (x) = \frac{x^{3} + 6}{x^{2}}

See what the power rule cannot do

The power rule handles one power of $x$ , and this is a quotient. Differentiating the top and the bottom separately would give $3 x^{2} / (2 x)$ , which is wrong. Rewrite the quotient as a sum of powers first. The function is defined for $x \neq 0$ .

f (x) = \frac{x^{3} + 6}{x^{2}}, x \neq 0

Divide each term by $x^{2}$

$x^{3} / x^{2} = x$ and $6 / x^{2} = 6 x^{- 2}$ . Dividing term by term is allowed because the denominator is a single term.

f (x) = \frac{x^{3}}{x^{2}} + \frac{6}{x^{2}} = x + 6 x^{- 2}

\begin{aligned} f (x) & = \frac{x^{3}}{x^{2}} + \frac{6}{x^{2}} \\ = x + 6 x^{- 2} \end{aligned}

Apply the power rule term by term

$x$ has derivative $1$ . For $6 x^{- 2}$ , bring down $- 2$ and lower the exponent to $- 3$ , which gives $6 \cdot (- 2) x^{- 3} = - 12 x^{- 3}$ .

f^{'} (x) = 1 - 12 x^{- 3}

Write the answer without negative exponents

$x^{- 3}$ is $1 / x^{3}$ , so the derivative is $1 - 12 / x^{3}$ , still for $x \neq 0$ .

f^{'} (x) = 1 - \frac{12}{x^{3}}

Find the slope at $x = 2$

$2^{3} = 8$ , so $f^{'} (2)$ $= 1 - 12 / 8$ $= 1 - 1.5$ $= - 0.5$ . The graph is falling at $x = 2$ .

f^{'} (2) = 1 - \frac{12}{8} = - \frac{1}{2}

Test the wrong method

The shortcut $3 x^{2} / (2 x)$ simplifies to $1.5 x$ , which is $3$ at $x = 2$ and says the graph is rising. A secant settles it. $f (2) = 14 / 4 = 3.5$ and $f (2.01) \approx 3.49511$ , a slope of about $- 0.489$ . The graph falls, and $- 1 / 2$ is right.

Result

$f^{'} (x) = 1 - 12 / x^{3}$ for $x \neq 0$ , and $f^{'} (2) = - 1 / 2$ .

f^{'} (x) = 1 - \frac{12}{x^{3}}

Your turn

1. One like the example

Differentiate $g (x) = (2 x^{3} - 5) / x$ , then find $g^{'} (1)$ .

Show the answer and explanation

$g^{'} (x) = 4 x + 5 / x^{2}$ , and $g^{'} (1) = 9$ .

Divide each term by $x$ to get $2 x^{2} - 5 x^{- 1}$ . The power rule gives $4 x$ and $- 5 \cdot (- 1) x^{- 2} = 5 x^{- 2}$ . At $x = 1$ , $g^{'} (1) = 4 + 5 = 9$ .

\begin{array}{l} g (x) = \frac{2 x^{3} - 5}{x} \\ g (x) = 2 x^{2} - 5 x^{- 1} \\ g^{'} (x) = 4 x + 5 x^{- 2} \\ g^{'} (1) = 9 \end{array}

2. A variation

Differentiate $f (x) = (x^{2} + 1) (2 x - 3)$ . This time the function is a product.

Show the answer and explanation

$f^{'} (x) = 6 x^{2} - 6 x + 2$ .

Multiplying out gives $2 x^{3} - 3 x^{2} + 2 x - 3$ , a sum of powers. Term by term, the derivative is $6 x^{2} - 6 x + 2$ . Multiplying the factors’ derivatives, $2 x \cdot 2 = 4 x$ , would be wrong.

\begin{array}{l} f (x) = (x^{2} + 1) (2 x - 3) \\ f (x) = 2 x^{3} - 3 x^{2} + 2 x - 3 \\ f^{'} (x) = 6 x^{2} - 6 x + 2 \end{array}

3. Somewhere new

A balloon is a sphere of radius $r$ $cm$ , with volume $V (r) = (4 / 3) π r^{3}$ ${cm}^{3}$ . How fast does the volume grow with the radius when $r = 3 cm$ ?

Show the answer and explanation

$V^{'} (3) = 36 π \approx 113 {cm}^{3}$ per $cm$ .

The constant $4 / 3 \cdot π$ stays in front, and the power rule turns $r^{3}$ into $3 r^{2}$ . $V^{'} (r) = 4 π r^{2}$ , which is the sphere’s surface area. At $r = 3$ , $V^{'} (3)$ $= 4 π \cdot 9$ $= 36 π$ $\approx 113.1 {cm}^{3}$ for each extra centimeter of radius.

\begin{array}{l} V (r) = \frac{4}{3} π r^{3} \\ V^{'} (r) = 4 π r^{2} \\ V^{'} (3) = 36 π \end{array}

Keep exploring

Open the steps in Math: the Checker marks every row ✓. Write $f^{'} (x) = 3 x^{2} / (2 x)$ under the first row instead and it reads ✗, because a quotient’s derivative is not the quotient of the derivatives.

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See what the power rule cannot do

Divide each term by x2x2

Apply the power rule term by term

Write the answer without negative exponents

Find the slope at x=2x=2

Test the wrong method

Result

Your turn

1. One like the example

g′(x)=4⁢x+5⁢/x2g′(x)=4x+5/x2, and g′(1)=9g′(1)=9.

2. A variation

f′(x)=6x2−6⁢x+2f′(x)=6x2−6x+2.

3. Somewhere new

V′(3)=36⁢π≈113 cm3V′(3)=36π≈113 cm3 per cmcm.

Keep exploring

Study it, then try it

Divide each term by $x^{2}$

Find the slope at $x = 2$

$g^{'} (x) = 4 x + 5 / x^{2}$ , and $g^{'} (1) = 9$ .

$f^{'} (x) = 6 x^{2} - 6 x + 2$ .

$V^{'} (3) = 36 π \approx 113 {cm}^{3}$ per $cm$ .