3.3 Differentiation rules (Page 5/14)

Page 5 / 14

\frac{d}{d} (x^{− n}) = \frac{0 (x^{n}) - 1 (n x^{n - 1})}{{(x^{n})}^{2}} .

Simplifying, we see that

\frac{d}{d} (x^{− n}) = \frac{− n x^{n - 1}}{x^{2 n}} = − n x^{(n - 1) - 2 n} = − n x^{− n - 1} .

Finally, observe that since $k = − n,$ by substituting we have

\frac{d}{d x} (x^{k}) = k x^{k - 1} .

□

Using the extended power rule

Find $\frac{d}{d x} (x^{−4}) .$

By applying the extended power rule with $k = −4,$ we obtain

\frac{d}{d x} (x^{−4}) = −4 x^{−4 - 1} = −4 x^{−5} .

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Using the extended power rule and the constant multiple rule

Use the extended power rule and the constant multiple rule to find $f (x) = \frac{6}{x^{2}} .$

It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as $f (x) = 6 x^{−2} .$

\begin{matrix} f^{'} (x) & = \frac{d}{d x} (\frac{6}{x^{2}}) = \frac{d}{d x} (6 x^{−2}) & Rewrite \frac{6}{x^{2}} as 6 x^{−2} . \\ = 6 \frac{d}{d x} (x^{−2}) & Apply the constant multiple rule. \\ = 6 (−2 x^{−3}) & Use the extended power rule to differentiate x^{−2} . \\ = −12 x^{−3} & Simplify. \end{matrix}

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Find the derivative of $g (x) = \frac{1}{x^{7}}$ using the extended power rule.

$g^{'} (x) = −7 x^{−8} .$

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Combining differentiation rules

As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.

Combining differentiation rules

For $k (x) = 3 h (x) + x^{2} g (x),$ find $k^{'} (x) .$

Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.

\begin{matrix} k^{'} (x) & = \frac{d}{d x} (3 h (x) + x^{2} g (x)) = \frac{d}{d x} (3 h (x)) + \frac{d}{d x} (x^{2} g (x)) & Apply the sum rule. \\ = 3 \frac{d}{d x} (h (x)) + (\frac{d}{d x} (x^{2}) g (x) + \frac{d}{d x} (g (x)) x^{2}) & \begin{matrix} Apply the constant multiple rule to \\ differentiate 3 h (x) and the product \\ rule to differentiate x^{2} g (x) . \end{matrix} \\ = 3 h^{'} (x) + 2 x g (x) + g^{'} (x) x^{2} \end{matrix}

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Extending the product rule

For $k (x) = f (x) g (x) h (x),$ express $k^{'} (x)$ in terms of $f (x), g (x), h (x),$ and their derivatives.

We can think of the function $k (x)$ as the product of the function $f (x) g (x)$ and the function $h (x) .$ That is, $k (x) = (f (x) g (x)) \cdot h (x) .$ Thus,

\begin{matrix} k^{'} (x) & = \frac{d}{d x} (f (x) g (x)) \cdot h (x) + \frac{d}{d x} (h (x)) \cdot (f (x) g (x)) & \begin{matrix} Apply the product rule to the product \\ of f (x) g (x) and h (x) . \end{matrix} \\ = (f^{'} (x) g (x) + g^{'} (x) f (x) h) (x) + h^{'} (x) f (x) g (x) & Apply the product rule to f (x) g (x) . \\ = f^{'} (x) g (x) h (x) + f (x) g^{'} (x) h (x) + f (x) g (x) h^{'} (x). & Simplify. \end{matrix}

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Combining the quotient rule and the product rule

For $h (x) = \frac{2 x^{3} k (x)}{3 x + 2},$ find $h^{'} (x) .$

This procedure is typical for finding the derivative of a rational function.

\begin{matrix} h^{'} (x) & = \frac{\frac{d}{d x} (2 x^{3} k (x)) \cdot (3 x + 2) - \frac{d}{d x} (3 x + 2) \cdot (2 x^{3} k (x))}{{(3 x + 2)}^{2}} & Apply the quotient rule. \\ = \frac{(6 x^{2} k (x) + k^{'} (x) \cdot 2 x^{3}) (3 x + 2) - 3 (2 x^{3} k (x))}{{(3 x + 2)}^{2}} & \begin{matrix} Apply the product rule to find \\ \frac{d}{d x} (2 x^{3} k (x)) . Use \frac{d}{d x} (3 x + 2) = 3 . \end{matrix} \\ = \frac{−6 x^{3} k (x) + 18 x^{3} k (x) + 12 x^{2} k (x) + 6 x^{4} k^{'} (x) + 4 x^{3} k^{'} (x)}{{(3 x + 2)}^{2}} & Simplify. \end{matrix}

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Find $\frac{d}{d x} (3 f (x) - 2 g (x)) .$

$3 f^{'} (x) - 2 g^{'} (x) .$

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Determining where a function has a horizontal tangent

Determine the values of $x$ for which $f (x) = x^{3} - 7 x^{2} + 8 x + 1$ has a horizontal tangent line.

To find the values of $x$ for which $f (x)$ has a horizontal tangent line, we must solve $f^{'} (x) = 0 .$ Since

f^{'} (x) = 3 x^{2} - 14 x + 8 = (3 x - 2) (x - 4),

we must solve $(3 x - 2) (x - 4) = 0 .$ Thus we see that the function has horizontal tangent lines at $x = \frac{2}{3}$ and $x = 4$ as shown in the following graph.

The graph shows f(x) = x3 – 7x2 + 8x + 1, and the tangent lines are shown as x = 2/3 and x = 4. — This function has horizontal tangent lines at x = 2/3 and x = 4.

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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