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Section 8.6 Derivatives Take Practice

¶I want to recommend that you practice as much as possible. You might find it useful to do some of this practice using the following web-based app that will also work on smart phones or tablets: Derivative Practice on Algebraic Formulas. Work your way until you can do all of the types of calculations without hesitation.

Start by knowing basic derivatives of power functions using the power rule

\begin{equation*}
\frac{d}{dx}[x^p] = p x^{p-1}.
\end{equation*}

We looked at why this rule is true when \(p\) is a positive integer, but the rule is true for any power function. Combining this with the constant multiple rule, you can find the derivative

\begin{equation*}
\frac{d}{dx}[A x^p] = Ap x^{p-1}.
\end{equation*}

######
Example 8.6.1

Compute the following derivatives:

- \(\displaystyle \frac{d}{dx}[5x^3]\)
- \(\displaystyle \frac{d}{dx}[\frac{x^4}{7}]\)
- \(\displaystyle \frac{d}{dx}[\frac{2}{7x^2}]\)

Solution
- To compute \(\displaystyle \frac{d}{dx}[5x^3]\text{,}\) we recognize the elementary power \(x^3\) which has power \(p=3\) so that its derivative is \(\frac{d}{dx}[x^3] = 3x^2\text{.}\) Use the constant multiple rule to get the final derivative.
\begin{equation*}
\frac{d}{dx}[5x^3] = 5 (3x^2) = 15 x^2.
\end{equation*}

- To compute \(\displaystyle \frac{d}{dx}[\frac{x^4}{7}]\text{,}\) we recognize the elementary power \(x^4\) which has power \(p=4\) so that its derivative is \(\frac{d}{dx}[x^4] = 4x^3\text{.}\) The fraction is a constant multiple in disguise with constant \(\frac{1}{7}\text{.}\)
\begin{equation*}
\frac{d}{dx}[\frac{x^4}{7}] = \frac{d}{dx}[\frac{1}{7} x^4] = \frac{1}{7} (4x^3) = \frac{4}{7} x^3.
\end{equation*}

- To compute \(\displaystyle \frac{d}{dx}[\frac{2}{7x^2}]\text{,}\) we use the properties of powers to rewrite division by a power as a negative power,
\begin{equation*}
\frac{2}{7x^2} = \frac{2}{7} x^{-2}.
\end{equation*}

The basic power \(p=-2\) has a derivative with new power \(p-1=-3\text{,}\) so
\begin{equation*}
\frac{d}{dx}[ \frac{2}{7x^2} ] = \frac{d}{dx}[ \frac{2}{7} x^{-2} ]
= \frac{2}{7} (-2x^{-3}) = \frac{-4}{7x^3}.
\end{equation*}

Once you have mastered these elementary building blocks with the constant multiple rule, you can move to sums of these building blocks. Derivatives behave nicely with sums, since the derivative of a sum is the sum of the derivatives,

\begin{equation*}
\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)] = f'(x) + g'(x).
\end{equation*}

In practice, this means that as soon as you recognize a function is combined as a sum of elementary parts, you can just compute the derivatives of each part separately and add the results. (Subtraction is just addition with an inverse, so both are done at the same time.)

######
Example 8.6.2

Compute the following derivatives:

- \(\displaystyle \frac{d}{dx}[2x^5-3x^3+5x^2+7]\)
- \(\displaystyle \frac{d}{dx}[x^2 + 3x - \frac{1}{2x^3}]\)

Solution
For each problem, pay attention to how the differentiation operator is applied, starting from the entire formula to individual components until the ultimate answer is found.

\begin{align*}
\frac{d}{dx}[2x^5-3x^3+5x^2+7] &= \frac{d}{dx}[2x^5] + \frac{d}{dx}[-3x^3] + \frac{d}{dx}[5x^2] + \frac{d}{dx}[7]\\
&= 2(5x^4) + -3(3x^2) + 5(2x) + 0 \\
& = 10x^4 -9x^2 + 10x
\end{align*}

\begin{align*}
\displaystyle \frac{d}{dx}[x^2 + 3x - \frac{1}{2x^3}]
&= \frac{d}{dx}[x^2] + \frac{d}{dx}[3x] + \frac{d}{dx}[-\frac{1}{2}x^{-3}] \\
&= 2x + 3 + \frac{-1}{2}(-3x^{-4}) \\
&= 2x+3+\frac{3}{2x^4}
\end{align*}

Now that you can compute derivatives of sums of elementary terms, you should practice computing derivatives of products. The product rule for derivatives do not follow the same simple rule as sums. A little memorization jingle that might help is, "The derivative of \(u\) times \(v\) is ue-dee-vee plus vee-dee-ue," which as formula is

\begin{equation*}
\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}.
\end{equation*}

Alternatively, I personally use a tactile approach where I touch each factor one at a time and write down a new product where I replace the factor I am touching with its derivative and leave all other factors alone, adding the results. For a product of \(u\) and \(v\text{,}\) I would write

\begin{equation*}
\frac{d}{dx}[ u \cdot v] = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx},
\end{equation*}

and for a product of three terms, \(u, v, w\text{,}\) I would write

\begin{equation*}
\frac{d}{dx}[u\cdot v \cdot w] = \frac{du}{dx} \cdot v \cdot w
+ u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}.
\end{equation*}

######
Example 8.6.3

Compute the following derivatives:

- \(\displaystyle \frac{d}{dx}[(2x+5)(3x-7)]\)
- \(\displaystyle \frac{d}{dx}[x^2(x^3+5)]\)
- \(\displaystyle \frac{d}{dx}[4x^2(3x-1)(4x+5)]\)

Solution
For each problem, continue to watch how the differentiation operator is applied, starting from the entire formula to individual components until the ultimate answer is found.

\begin{align*}
\frac{d}{dx}[(2x+5)(3x-7)] &= \frac{d}{dx}[2x+5] \cdot (3x-7) + (2x+5) \cdot [3x-7] \\
&= 2(3x-7) + (2x+5)(3) \\
&= 6x-14 + 6x+15 \\
&= 12x+1
\end{align*}

\begin{align*}
\frac{d}{dx}[x^2(x^3+5)] &= \frac{d}{dx}[x^2] \cdot (x^3+5) + x^2 \cdot \frac{d}{dx}[x^3+5] \\
&= (2x)(x^3+5) + x^2(3x^2+0) \\
&= 2x^4+10x + 3x^4 \\
&= 5x^4+10x
\end{align*}

\begin{align*}
\frac{d}{dx}[4x^2(3x-1)(4x+5)] &= \frac{d}{dx}[4x^2] (3x-1)(4x+5) \\
& \qquad + (4x^2) \frac{d}{dx}[3x-1] (4x+5) \\
& \qquad + (4x^2)(3x-1) \frac{d}{dx}[4x+5] \\
&= (8x)(3x-1)(4x+5) + (4x^2)(3)(4x+5) + (4x^2)(3x-1)(4) \\
&= 8x(12x^2+15x-4x-5) + 12x^2(4x+5) + 16x^2(3x-1) \\
&= 96x^3 + 88x^2-40x + 48x^3+60x^2 +48x^3-16x^2 \\
&= 192x^3 +132x^2-40x
\end{align*}

In each of these examples, it would also be possible to multiply out the formulas before taking a derivative. This is often easier because then you only need to use the sum rule rather than the product rule.

\begin{align*}
\frac{d}{dx}[(2x+5)(3x-7)] &= \frac{d}{dx}[6x^2-14x+15x-35] \\
&= \frac{d}{dx}[6x^2+x-35] \\
&= 12x+1
\end{align*}

\begin{align*}
\frac{d}{dx}[x^2(x^3+5)] &= \frac{d}{dx}[x^5+5x^2] \\
&= 5x^4+10x
\end{align*}

\begin{align*}
\frac{d}{dx}[4x^2(3x-1)(4x+5)] &= \frac{d}{dx}[4x^2(12x^2+15x-4x-5)] \\
&= \frac{d}{dx}[48x^4+44x^3-20x^2] \\
&= 48(4x^3) +44(3x^2)-20(2x)\\
&= 192x^3 +132x^2-40x
\end{align*}

However, it is good to practice the product rule for those cases later where it is not possible to expand a formula so that the product rule isn't necessary.

After the product rule, you should master the quotient rule,

\begin{equation*}
\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}.
\end{equation*}

A memory phrase to remember this rule is "When the derivative of a quotient you want to know, it's low-dee-high minus high-dee-low. But don't forget it's over low-low." In symbols, this rhyme would be written

\begin{equation*}
\frac{d}{dx}[\frac{\hbox{Hi}}{\hbox{Lo}}] = \frac{\hbox{Lo} \frac{d}{dx}[\hbox{Hi}] - \hbox{Hi} \frac{d}{dx}[\hbox{Lo}]}{\hbox{Lo}\hbox{Lo}}.
\end{equation*}

######
Example 8.6.4

Compute the following derivatives:

- \(\displaystyle \frac{d}{dx}\left[\frac{2x+5}{3x-7}\right]\)
- \(\displaystyle \frac{d}{dx}\left[\frac{x^2}{x^3+5}\right]\)

Solution
Applying the quotient rule for derivatives leads to each answer. You do not need to expand the square of the denominator, but you should simplify the numerator.

\begin{align*}
\frac{d}{dx}\left[\frac{2x+5}{3x-7}\right]
&= \frac{(3x-7) \frac{d}{dx}[2x+5] - (2x+5)\frac{d}{dx}[3x-7]}{(3x-7)^2} \\
&= \frac{(3x-7)(2) - (2x+5)(3)}{(3x-7)^2} \\
&= \frac{6x-14-6x-15}{(3x-7)^2} \\
&= \frac{-29}{(3x-7)^2}
\end{align*}

\begin{align*}
\frac{d}{dx}\left[\frac{x^2}{x^3+5}\right]
&= \frac{(x^3+5)\frac{d}{dx}[x^2] -x^2 \frac{d}{dx}[x^3+5]}{(x^3+5)^2} \\
&= \frac{(x^3+5)(2x) - x^2(3x^2)}{(x^3+5)^2} \\
&= \frac{2x^4+10x-3x^4}{(x^3+5)^2} \\
&= \frac{-x^4+10x}{(x^3+5)^2}
\end{align*}