Welcome back to AP Calculus with Fiveable! Now that you’ve mastered finding derivatives of trigonometric functions sinx and cosx, let’s cover the rest! Remembering these rules is key to simplifying your calculus journey. 🌟
First, let’s have a glance at a summary table for quick reference.
Function | Derivative |
Tangent Function: f(x)=tanx | f′(x)=sec2x |
Cotangent Function: g(x)=cotx | g′(x)=−csc2x |
Secant Function: h(x)=secx | h′(x)=secxtanx |
Cosecant Function: k(x)=cscx | k′(x)=−cscxcotx |
It's important to note that these are only valid for angles in radians, not degrees.
The derivative of tanx is sec2x. Let’s consider an example:
f(x)=3tanx+2x2
To find the derivative of this equation, differentiate 3tanx and 2x2 individually.
Since the derivative of tanx is sec2x, the derivative of the first part is 3sec2x. The derivative of 2x2 is 4x. Hence, f′(x)=3sec2x+4x.
The derivative of cotx is −csc2x. For example:
f(x)=5cotx+x
We again have to differentiate the two terms separately! The derivative of cotx is −csc2x, so the derivative of the first term is −5csc2x. The derivative of x is 1. Therefore, f′(x)=−5csc2x+1 or f′(x)=1−5csc2x.
The derivative of secx is secxtanx. As an example:
f(x)=2secx+3x3
Knowing the above trig derivative rule, the derivative of the first term is 2secxtanx. The derivative of 3x3 is 9x2. Thus, f′(x)=2secxtanx+9x2.
Last but not least, the derivative of cscx is −cscxcotx. For instance:
f(x)=4cscx+7x2
The derivative of the first part is −4cscxcotx. The derivative of 7x2 is 14x. Therefore, f′(x)=−4cscxcotx+14x.
Here are a couple of questions for you to get the concepts down!
Find the derivatives for the following problems.
- f(x)=2tan(x)+sec(x)
- f(x)=csc(x)cot(x)
- g(x)=tan2(6x)
- h(x)=5cot(x)
💡 Before we reveal the answers, remember to use the chain rule, sum rule, and quotient rules.
- f′(x)=2sec2(x)+sec(x)tan(x)
- f′(x)=−csc2(x)
- g′(x)=2tan(6x)(cos2(6x)1)
- h′(x)=−5csc2(x)
Practice these rules, and you’ll soon find them as intuitive as the basic derivatives! Keep up the great work. 🌈