Derivative Rules

In what follows, f and g are differentiable functions of x and k and n are constants.

1. Constant Multiple Rule: [latex] \frac{d}{dx} (kf) = kf\prime [/latex]
2. Sum (or Difference) Rule: [latex] \frac{d}{dx}(f+g)=f\prime + g\prime [/latex] or [latex] \frac{d}{dx}(f-g)=f\prime - g\prime [/latex]
3. Power Rule: [latex] \frac{d}{dx}(x^n) = nx^{n-1} [/latex] Special cases: [latex] \frac{d}{dx}(k) = 0 [/latex] (because k = kx⁰) and [latex] \frac{d}{dx}(x) = 1 [/latex] (because x = x¹)
4. Exponential Functions: [latex] \frac{d}{dx}(e^x) = e^x [/latex] [latex] \frac{d}{dx}(a^x) = \text{ln} a \times a^x [/latex]
5. Natural Logarithm: [latex] \frac{d}{dx}(\text{ln}x) = \frac{1}{x} [/latex]

Antiderivative Rules: Building Blocks

In what follows, f and g are differentiable functions of x and k, n, and C are constants.

1. Constant Multiple Rule: [latex] \int kf(x)dx = k \int f(x)dx [/latex]
2. Sum (or Difference) Rule: [latex] \int f(x) \pm g(x)dx = \int f(x)dx \pm \int f(x)dx [/latex]
3. Power Rule: [latex] \int x^n dx = \frac{x^{n+1}}{n + 1} + C [/latex] provided that n = −1 Special case: [latex] \int kdx = kx + C [/latex] (because k = kx⁰)
4. Exponential Functions: [latex] \int e^x dx = e^x + C [/latex] [latex] \int a^x dx = \frac{a^x}{\text{ln}a} + C [/latex]
5. Natural Logarithm: [latex] \int x^{-1}dx = \int\frac{1}{x}dx = \text{ln}|x| + C [/latex]

Example

Find the antiderivative of [latex] 3x^7 - 15\sqrt x + \frac{14}{x^2} [/latex]

Solution

[latex-display] \int(3x^7 - 15\sqrt x + \frac{14}{x^2})dx = \int(3x^7 - 15^{1/2} + 14x^{-2})dx = 3\frac{x^8}{8} - 15\frac{x^{3/2}}{3/2} + 14\frac{x^{-1}}{-1} + C [/latex-display] That’s a little hard to look at, so you might want to simplify a little: [latex-display] \int(3x^7 - 15\sqrt x + \frac{14}{x^2})dx = \frac{3x^8}{8} -10x^{3/2} - 14x^{-1} + C [/latex-display]

Example

[latex-display] \int(e^x + 12 - \frac{16}{x})dx [/latex-display]

Solution

[latex-display] \int(e^x + 12 - \frac{16}{x})dx = e^x +12x - 16\mathrm{ln}|x| + C [/latex-display]

Example

Find F(x) so that F′(x) = e^x and F(0) = 10.

Solution

This time we are looking for a particular antiderivative; we need to find exactly the right constant. Let’s start by finding the antiderivative: [latex-display] \int e^xdx=e^x + C [/latex-display] So we know that F′(x) = e^x +  some constant; we just need to find which one. For that, we’ll use the other piece of information (the initial condition): [latex-display] F(x) = e^x + C [/latex-display] [latex-display] F(0) = e^0 + C = 1 + C = 10 [/latex-display] [latex-display] C = 9 [/latex-display] The particular constant we need is 9; F(x) = e^x + 9. The reason we are looking at antiderivatives right now is so we can evaluate definite integrals exactly. Recall the Fundamental Theorem of Calculus: [latex-display] \int_{a}^{b} F\prime(x)dx = F(b) -F(a) [/latex-display] If we can find an antiderivative for the integrand, we can use that to evaluate the definite integral. The evaluation F(b) − F(a) is represented by the notation [latex] \left.F(x)\right]_a^b [/latex] or [latex] \left.F(x)\right|_a^b [/latex].

Example

Evaluate [latex] \int_{1}^{3} xdx [/latex] in two ways:

1. By sketching the graph of y = x and geometrically finding the area.
2. By finding an antiderivative of F(x) of the integrand and evaluating F(3) – F(1).

Solution

1. The graph of y = x is shown in figure 1, and the shaded region has area 4.
   Figure 1
2. One antiderivative of x is [latex] F(x) = \frac{1}{2}x^2 [/latex] (check by differentiating ), and [latex] \int_{1}^{3}xdx=\frac{1}{2}x^2]_1^3 = [\frac{1}{2}(3)^2] - [\frac{1}{2}(1)^2] = \frac{9}{2} - \frac{1}{2} = 4 [/latex]. Note that this answer agrees with the answer we got geometrically.

If we had used another antiderivative of x, say [latex] F(x) = \frac{1}{2}x^2 + 7 [/latex] (check by differentiating ), then [latex] F(x)|_{1}^{3} = F(3) - F(1) = [\frac{1}{2}(3^2)+7]-[\frac{1}{2}(1^2)+7]= \frac{23}{2} - \frac{15}{2} = 4 [/latex]. Whatever constant you choose, it gets subtracted away during the evaluation; we might as well always choose the easiest one, where the constant = 0.

Example

Find the area between the graph of y = 3x² and the horizontal axis for x between 1 and 2.

Solution

This is [latex] \int_{1}^{2}3x^2dx=x^3]_{1}^{2} = (2^3) -(1^3) = 7 [/latex].

Example

A robot has been programmed so that when it starts to move, its velocity after t seconds will be 3t² feet/second.

1. How far will the robot travel during its first 4 seconds of movement?
2. How far will the robot travel during its next 4 seconds of movement?

Solution

1. The distance during the first 4 seconds will be the area under the graph (figure 2) of velocity, from t = 0 to t = 4.
   Figure 2
   That area is the definite integral [latex] \int_{0}^{4}3t^2dt [/latex]. An antiderivative of 3t² is t³, so [latex] \int_{0}^{4}3t^2dt =t^3 ]_{0}^{4} = 4^3 - 0^3 =64 [/latex] feet.
2. [latex] \int_{4}^{8}3t^2dt =t^3 ]_{4}^{8} = 8^3 - 4^3 = 512 - 64 = 448 [/latex] feet.

Example

Suppose that t minutes after putting 1000 bacteria on a Petri plate the rate of growth of the population is 6t bacteria per minute.

1. How many new bacteria are added to the population during the first 7 minutes?
2. What is the total population after 7 minutes?

Solution

1. The number of new bacteria is the area under the rate of growth graph (figure 3), and one antiderivative of 6t is 3t².
   Figure 3
   So new bacteria = [latex] \int_{0}^{7} 6t \mathrm{d}t = 3t^2|_{0}^{7} = 3(7)^2 - 3(0)^2 = 147 [/latex].
2. The new population = {old population} + {new bacteria} = 1000 + 147 = 1147 bacteria.