Derivatives of powers, sums, and constant multiples
Power rule
The
power rule tells us how to take the derivative of any power of $x$:
If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.
This rule works for any constant power $n$. In differential notation, this reads
$\dfrac{d}{dx}[x^n] = nx^{n-1}$.
Sum and constant multiple rules
If $f'(x)$ and $g'(x)$ exist, and $c$ is constant, then
| $[f(x) \pm g(x)]' = f'(x) \pm g'(x) \qquad$ |
| $[cf(x)]' = cf'(x) \qquad$ |
In words
Examples: Derivatives of powers, sums, and constant multiples
| $\displaystyle \frac{d}{dx}[x^{3.2}]$ | ${}=3.2x^{2.2}$ |
| |
| $\displaystyle \frac{d}{dx}[4x^{-1}]$ | $\displaystyle {}=4\frac{d}{dx}[x^{-1}]$ |
| ${}= 4(-1)x^{-2}$ |
| ${}= -4x^{-2}$ |
|
| |
| $\displaystyle \frac{d}{dx}[3x^{1.1}+4x^{-2}]$ | $\displaystyle {}=\frac{d}{dx}[3x^{1.1}]+\frac{d}{dx}[4x^{-2}]$ |
| $\displaystyle {}=3\frac{d}{dx}[x^{1.1}]+4\frac{d}{dx}[x^{-2}]$ |
| ${}= 3(1.1)x^{0.1}+4(-2)x^{-3}$ |
| ${}= 3.3x^{0.1}-8x^{-3}$ |
|
| |
| $\displaystyle \frac{d}{dx}\left[\frac{1}{x}\right]$ | $\displaystyle {}=\frac{d}{dx}[x^{-1}]$ |
| $\displaystyle {}=(-1)x^{-2}$ |
| $\displaystyle {}= -\frac{1}{x^2}$ |
| |
| $\displaystyle \frac{d}{dx}\left[\sqrt{x}\right]$ | $\displaystyle {}=\frac{d}{dx}[x^{1/2}]$ |
| $\displaystyle {}=\frac{1}{2}x^{-1/2}$ |
| $\displaystyle {}= \frac{1}{2\sqrt{x}}$ |
Practice:
Practice:
Want more practice? Try the
online tutorial or the
online review exercises.
Application: Marginal analysis
If $Q(x)$ represents any quantity such as cost, revenue, profit or loss on the sale of $x$ items, then $Q'(x)$ is called the
marginal quantity. Thus, for instance, the marginal cost measures the increase in total cost per item. This is effectively the cost of each additional item.
The marginal cost is distinct from the
average cost, which measures the average of the total cost of the first $x$ items. Average cost is given by
| $\bar{C}$ | $\displaystyle {}=\frac{C(x)}{x}$ |
| $\displaystyle {}=\frac{\text{Total cost}}{\text{Number of items}}$ |
Visualizing average and marginal cost
Examples: Application: Marginal analysis
Suppose the cost of the first $x$ items is given by
$C(x) = 4x^{0.2}- 0.1x$ €.
Then the marginal cost is
$C'(x) = 0.8x^{-0.8}- 0.1$ € per item.
In particular, $C'(3) = 0.8(3)^{-0.8}- 0.1 \approx 0.23$ €/item is the approximate cost of the fourth item. On the other hand,
$\displaystyle \bar{C}(x) = \frac{C(x)}{x} = \frac{4x^{0.2}- 0.1x}{x} = 4x^{-0.8}- 0.1$ € per item.
In particular, $\bar{C}(3) = 4(3)^{-0.8} - 0.1 \approx 1.56$ €/item is the average cost of the first three items.
Practice:
Derivatives of products and quotients
Product rule
If $f$ and $g$ are differentiable functions of $x$, then so is their product $fg$, and
$\displaystyle \frac{d}{dx}\left[f(x)g(x)\right] = \color{#ff504d}{f'(x)}g(x) + f(x)\color{#ff504d}{g'(x)}.$
Product rule in words
Quotient rule
If $f$ and $g$ are differentiable functions of $x$, then so is their quotient $f/g$, and
$\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\color{#ff504d}{f'(x)}g(x) - f(x)\color{#ff504d}{g'(x)}}{[g(x)^2]}.$
Quotient rule in words
Examples: Derivatives of products and quotients
$\displaystyle \frac{d}{dx}\left[x^2(3x-1)\right] = \color{#ff504d}{2x}(3x-1) + x^2(\color{#ff504d}{3}) \qquad$
(The derivatives of $f$ and $g$ are shown
in color.)
$\displaystyle \frac{d}{dx}\left[\frac{x^3}{x^2+1}\right] = \frac{\color{#ff504d}{3x^2}(x^2+1) + x^3(\color{#ff504d}{2x})}{(x^2+1)^2} \qquad$
Of course, you should simplify the answers and not leave them like that!
Practice:
Combining rules for differentiation: Calculation thought experiment
The
calculation thought experiment (CTE) is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, or difference. Given such an expression, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient, and so on.
Using the calculation thought experiment (CTE) to differentiate a function
If the CTE says, for instance, that the expression is a sum of two smaller expressions, then apply the rule for sums as a first step. This will leave you having to differentiate the smaller expressions, and you can use the CTE on these, and so on...
Examples: Combining rules for differentiation: Calculation thought experiment
1. $(3x^2- 4)(2x+1)$ can be calculated by first calculating the expressions in parentheses and then multiplying. Since the last step is
multiplication, we can treat the expression as a
product.
2. $\dfrac{2x-1}{x}$ can be calculated by first calculating the numerator and denominator separately, and then dividing one by the other. Since the last step is
division, we can treat the expression as a
quotient.
3. $(4x-1)(x+2) + x^2$ can be calculated by first calculating the product $(4x-1)(x+2)$, then calculating $x^2$, and finally adding the two answers. Since the last step is
addition, we can treat the expression as a
sum.
4. $(3x^2-1)^5$ can be calculated by first calculating the expression in parentheses, and then raising the answer to the fifth power. Since the last step is
raising to a power, we can treat the expression as a
power. Derivatives of powers of functions other than $x$ are in the next part of this summary.
Practice:
Warning: sizeof(): Parameter must be an array or an object that implements Countable in
/home/zweigmedia2/www/www/summariesContent/summariesScriptsNV.php on line
51