Introduction
A sinusoidal model can oscillate slowly or rapidly depending on the coefficient multiplying its input variable. Differentiating the model reveals a second effect: that same coefficient scales the size of the derivative. A rapidly oscillating sinusoid therefore tends to have a larger derivative amplitude when its original amplitude is held fixed.
We will make both effects precise for sine and cosine models. The key calculation is the chain rule, but the goal is to understand what its factor means: how it changes the size of the derivative and how it relates to the rate at which the sinusoid completes its cycles.
Sinusoidal Models and Frequency Coefficients
Consider a sinusoidal model in the form
Here \(A\) is the signed amplitude parameter, \(\omega\) is the coefficient of \(x\) inside the trigonometric function, \(\phi\) is a phase shift, and \(D\) is a vertical shift. We take the angle in radians. The amplitude of the function, meaning its distance from its midline to a maximum, is \(|A|\). The coefficient \(\omega\) is called its angular frequency.
The sign of \(\omega\) affects the direction in which the input angle changes as \(x\) increases, but not how many complete cycles occur over a given interval. Thus, when describing the rate or period of oscillation, we use \(|\omega|\). The phase shift \(\phi\) changes where a cycle begins, and \(D\) moves the graph vertically; neither changes the oscillation rate.
How the Chain Rule Scales the Derivative
The derivative rules for sine and cosine are applied to the entire input \(\omega x+\phi\). The derivative of that input is \(\omega\), so the chain rule contributes a factor of \(\omega\).
Similarly, differentiating a cosine model gives
Since sine and cosine range between \(-1\) and \(1\), multiplying either by a constant \(C\) produces a function with amplitude \(|C|\). It follows that the derivative amplitude is \(|A\omega|=|A||\omega|\). This is the central scaling relationship: keeping the original amplitude fixed while multiplying the frequency coefficient by a factor multiplies the derivative amplitude by the same positive factor.
Why the Period Changes
A complete sine or cosine cycle occurs when its input angle changes by \(2\pi\). As \(x\) changes by an amount \(T\), the angle \(\omega x+\phi\) changes by \(\omega T\). For one complete cycle, the magnitude of this change must be \(2\pi\), so \(|\omega|T=2\pi\).
This is the period of the sinusoid, the positive amount of change in \(x\) after which its values repeat. The derivative contains sine or cosine with the same input \(\omega x+\phi\), so it repeats after this same amount \(T\). Its phase is different, but its period is unchanged.
These statements can also be seen directly from the derivative formulas. For the sine model, the derivative amplitude is \(|A||\omega|\). For the cosine model it is also \(|A||\omega|\), because the minus sign changes the direction of the derivative values but not their maximum magnitude. And the input coefficient remains \(\omega\) in both derivatives, so the cycle length remains \(2\pi/|\omega|\).
Read \(A\) and \(\omega\) from the model, including the coefficient of \(x\) inside the trigonometric function.
Apply the sine or cosine derivative rule and multiply by the derivative \(\omega\) of the inner expression.
The derivative amplitude is \(|A\omega|\), and the period of both functions is \(2\pi/|\omega|\), provided \(A\ne0\) and \(\omega\ne0\).
Worked Example: Comparing Two Sine Models
Worked Example: Comparing Two Sine Models
Compare the derivative amplitudes and periods of \(f(x)=2\sin(3x+\pi/4)\) and \(g(x)=2\sin(6x+\pi/4)\).
For \(f\), the original amplitude is \(2\), and the frequency coefficient is \(3\). By the chain rule,
Thus the derivative amplitude is \(|6|=6\). The period of \(f\), and also of \(f'\), is
For \(g\), the original amplitude is also \(2\), but its frequency coefficient is \(6\). Differentiating gives
Its derivative amplitude is \(|12|=12\), and its period is
Therefore, doubling the frequency coefficient from \(3\) to \(6\) doubles the derivative amplitude from \(6\) to \(12\), while halving the period from \(2\pi/3\) to \(\pi/3\). As a check, the derivatives have the required form \(A\omega\cos(\omega x+\phi)\), and substituting their coefficients into \(2\pi/|\omega|\) gives the stated periods.
Worked Example: A Cosine with a Negative Coefficient
Worked Example: A Cosine with a Negative Coefficient
Let \(p(x)=-4\cos\left(\frac{x}{2}-\frac{\pi}{3}\right)+7\). Find the derivative, its amplitude, and the period of both the model and its derivative.
Here \(A=-4\), \(\omega=\frac12\), and \(D=7\). Set \(u(x)=\frac{x}{2}-\frac{\pi}{3}\), so \(u'(x)=\frac12\). Using the chain rule,
The derivative amplitude is \(|2|=2\). This agrees with the scaling formula \(|A\omega|=|-4\cdot\frac12|=2\). The period is
The derivative has the same input coefficient \(\frac12\), so its period is also \(4\pi\). To check the sign and chain factor at \(x=0\), the input angle is \(-\pi/3\). The derivative formula gives \(p'(0)=2\sin(-\pi/3)=-\sqrt{3}\). Directly differentiating the original expression gives the same formula, including the positive coefficient \(2\). The vertical shift \(7\) contributes no derivative.
Angular Frequency and Cycles per Unit
Sometimes a model is written using the number of cycles per unit of \(x\), rather than the coefficient of \(x\) as an angular frequency. If \(r\) is the number of cycles per unit, then one cycle corresponds to an angle change of \(2\pi\). The model is written with input \(2\pi r x+\phi\), so its angular frequency is \(2\pi r\).
In this form, the period is \(1/|r|\) when \(r\ne0\), and the derivative amplitude is \(2\pi|r||A|\). The factor \(2\pi\) appears because \(r\) counts cycles, while the sine and cosine derivative rules use radian angle. Distinguishing these two conventions prevents a common mismatch between the frequency in cycles per unit and the coefficient inside the trigonometric function.
Worked Example: Using Cycles per Unit
Worked Example: Using Cycles per Unit
Consider \(q(x)=5\sin(2\pi\cdot 3x+\pi/6)\). Find its derivative amplitude and period.
The model completes \(3\) cycles per unit of \(x\), so its angular frequency is \(2\pi\cdot3=6\pi\). Differentiating the full input gives
The derivative amplitude is therefore \(|30\pi|=30\pi\). Its period is
This matches the cycles-per-unit description: three cycles occur in one unit, so each cycle has length \(1/3\). The derivative retains the same input and therefore has that same period. The derivative formula also confirms the amplitude: the original amplitude \(5\) is multiplied by the angular frequency \(6\pi\), giving \(30\pi\).
Common Mistake: Omitting the Frequency Factor
A frequent error is to differentiate the outer sine correctly but forget to multiply by the derivative of its input. For example, the calculation
is incorrect. The inner function \(5x\) has derivative \(5\), so the chain rule gives
The incorrect result would have derivative amplitude \(4\); the correct derivative amplitude is \(20\). The formula \(|A\omega|=|4\cdot5|=20\) provides a quick check. The correct derivative has period \(2\pi/5\), the same as the original function. Omitting the chain factor changes not only the derivative’s amplitude but also leaves the calculation inconsistent with the rate of change of the input angle.
When the Scaling Description Does Not Apply
The formulas for a nonconstant sinusoid assume both \(A\ne0\) and \(\omega\ne0\). If \(A=0\), the model is constant regardless of \(\omega\), and its derivative is zero. If \(\omega=0\), the input angle is constant, so the model is again constant and its derivative is zero. In either case there is no oscillation period to calculate for the resulting constant function.
For nonzero \(A\) and \(\omega\), keep the two effects separate: \(|A\omega|\) describes the derivative amplitude, while \(2\pi/|\omega|\) describes the period. Increasing \(|\omega|\) increases the derivative amplitude and decreases the period. Increasing \(|A|\) increases the derivative amplitude but does not change the period.
Check Your Understanding
For each question, identify the frequency coefficient inside the trigonometric function before applying the derivative rule.
- Find the derivative and derivative amplitude of \(h(x)=3\sin(4x-\pi/2)\).
- What is the period of \(h(x)=3\sin(4x-\pi/2)\), and why is it also the period of \(h'(x)\)?
- For \(k(x)=2\cos(7x+\pi/3)\), find the derivative and its amplitude.
- A model has amplitude \(6\) and angular frequency \(-2\). What is its derivative amplitude and period?
- Rewrite the angular frequency for a model completing \(5\) cycles per unit, and state the derivative-amplitude multiplier applied to the original amplitude.
