How do the derivatives of \(\tan(x)\text{,}\)\(\cot(x)\text{,}\)\(\sec(x)\text{,}\) and \(\csc(x)\) combine with other derivative rules we have developed to expand the library of functions we can quickly differentiate?
Because each angle \(\theta\) in standard position corresponds to one and only one point \((x,y)\) on the unit circle, the \(x\)- and \(y\)-coordinates of this point are each functions of \(\theta\text{.}\) In fact, this is the very definition of \(\cos(\theta)\) and \(\sin(\theta)\text{:}\)\(\cos(\theta)\) is the \(x\)-coordinate of the point on the unit circle corresponding to the angle \(\theta\text{,}\) and \(\sin(\theta)\) is the \(y\)-coordinate. From this simple definition, all of trigonometry is founded. For instance, the Fundamental Trigonometric Identity,
Because we know the derivatives of the sine and cosine function, we can now develop shortcut differentiation rules for the tangent, cotangent, secant, and cosecant functions. In this section’s preview activity, we work through the steps to find the derivative of \(y = \tan(x)\text{.}\)
Subsection2.4.2Derivatives of the cotangent, secant, and cosecant functions
In Preview Activity 2.4.1, we found that the derivative of the tangent function can be expressed in several ways, but most simply in terms of the secant function. Next, we develop the derivative of the cotangent function.
By the Fundamental Trigonometric Identity, we see that \(g'(x) = -\frac{1}{\sin^2(x)}\text{,}\) and recalling that \(\csc(x) = \frac{1}{\sin(x)}\text{,}\) it follows that we can express \(g'\) by the rule
Note that neither \(g\) nor \(g'\) is defined when \(\sin(x) = 0\text{,}\) which occurs at every integer multiple of \(\pi\text{.}\) Hence we have the following rule.
Notice that the derivative of the cotangent function is very similar to the derivative of the tangent function we discovered in Preview Activity 2.4.1.
Using the quotient rule we have determined the derivatives of the tangent, cotangent, secant, and cosecant functions, expanding our overall library of functions we can differentiate. Observe that just as the derivative of any polynomial function is a polynomial, and the derivative of any exponential function is another exponential function, so it is that the derivative of any basic trigonometric function is another function that consists of basic trigonometric functions. This makes sense because all trigonometric functions are periodic, and hence their derivatives will be periodic, too.
The derivative retains all of its fundamental meaning as an instantaneous rate of change and as the slope of the tangent line to the function under consideration.
Respond to each of the following prompts. Where a derivative is requested, be sure to label the derivative function with its name using proper notation.
When a mass hangs from a spring and is set in motion, the object’s position oscillates in a way that the size of the oscillations decrease. This is usually called a damped oscillation. Suppose that for a particular object, its displacement from equilibrium (where the object sits at rest) is modeled by the function
Assume that \(s\) is measured in inches and \(t\) in seconds. Sketch a graph of this function for \(t \ge 0\) to see how it represents the situation described. Then compute \(ds/dt\text{,}\) state the units on this function, and explain what it tells you about the object’s motion. Finally, compute and interpret \(s'(2)\text{.}\)
Each derivative exists and is defined on the same domain as the original function. For example, both the tangent function and its derivative are defined for all real numbers \(x\) such that \(x \ne \frac{k\pi}{2}\text{,}\) where \(k = \pm 1, \pm 2, \ldots\text{.}\)
The four rules for the derivatives of the tangent, cotangent, secant, and cosecant can be used along with the rules for power functions, exponential functions, and the sine and cosine, as well as the sum, constant multiple, product, and quotient rules, to quickly differentiate a wide range of different functions.
Find the equation of the tangent line to the curve \(y = 3\sec\mathopen{}\left(x\right)-6\cos\mathopen{}\left(x\right)\) at the point \(( \pi/3 , 3)\text{.}\) Write your answer in the form \(y = mx + b\) where \(m\) is the slope and \(b\) is the \(y\)-intercept.
An object moving vertically has its height at time \(t\) (measured in feet, with time in seconds) given by the function \(h(t) = 3 + \frac{2\cos(t)}{1.2^t}\text{.}\)
What is the object’s instantaneous velocity when \(t =2\text{?}\)
Explain why the function that you found in (a) is almost the opposite of the sine function, but not quite. (Hint: convert all of the trigonometric functions in (a) to sines and cosines, and work to simplify. Think carefully about the domain of \(f\) and the domain of \(f'\text{.}\))
