This page covers the first three basic rules when taking derivatives, the constant rule, constant multiple rule and the addition/subtraction rule.
Constant Rule
The constant rule is the simplest and most easily understood rule. The derivative calculates the slope, right? So, if you are given a horizontal line, what is the slope? Right! The slope is zero. That's it. That's the slope of every horizontal line. We can write the equation of a horizontal line as \(f(x)=c\) where \(c\) is a real number. Since these are always horizontal lines, the slope is zero. Therefore, the derivative of all constant functions (horizontal lines) is zero. We can derive this idea from the limit definition as follows. If \(f(x)=c\) \[ f~'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} = \lim_{h \to 0}{\frac{c - c}{h}} = \lim_{h \to 0}{0} = 0 \] Notice that c is gone from the final answer, \(f~'(x)=0\), so this holds for all horizontal lines. Thinking about this, it makes sense intuitively, right?
Constant Multiple Rule
This rule works as you would expect. Mathematically, it looks like this. \[ \frac{d}{dx}[cf(x)] = c \frac{d}{dx}[f(x)] \] Nothing surprising, just pull out the constant and take the derivative of the function. This is discussed in more detail with examples on the power rule page.
Addition and Subtraction Rules
When you have two functions that are added or subtracted, you just take the derivative of each individually. Mathematically, it looks like this.
\[ \frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)] \]
Nothing surprising or tricky here. It works just as you would expect.
[However, you will find out soon that this idea does NOT hold for multiplication and division. We have some special rules for those called the product rule and quotient rule.]
You will get plenty of chances to practice these techniques on the next page discussing the power rule.
You CAN Ace Calculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
---|---|---|---|
\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
To bookmark this page, log in to your account or set up a free account.
Single Variable Calculus |
---|
Multi-Variable Calculus |
---|
Differential Equations |
---|
Precalculus |
---|
Engineering |
---|
Circuits |
---|
Semiconductors |
---|
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
| |
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. |