Introduction
A sinusoidal model can be shifted horizontally without changing its basic sine or cosine form. For example, \(f(x)=4\sin(x-\pi/6)\) has a constant phase shift: the input to sine is \(x-\pi/6\), rather than \(x\). Differentiating such a model requires keeping track of both the outer trigonometric function and its shifted input.
The key point is that a constant phase shift stays inside the trigonometric function, while the derivative of the shifted input \(x-\phi\) is \(1\). The constant amplitude multiplies the derivative. These facts produce compact rules, but the minus sign in the cosine rule and the sign of the phase deserve careful attention.
Form of the models and conditions
Here \(A\) is fixed while \(x\) varies. In the usual amplitude convention, \(A\geq 0\); if a model is written with a negative coefficient, its sign can instead be handled as a separate multiplier. The phase constant may be positive, negative, or zero. For instance, \(x+\pi/4\) is the same form as \(x-\phi\) with \(\phi=-\pi/4\).
The results below assume the coefficient of \(x\) inside the trigonometric function is \(1\), and that the phase is constant. The derivative rules for sine and cosine require radians. If the phase or the input itself varies with \(x\), the derivative of the inner expression must also be included; the simple formulas here do not cover that different situation.
The derivative rules
In the sine formula, differentiation changes sine to cosine. In the cosine formula, differentiation changes cosine to negative sine. In both cases the amplitude coefficient is retained and the phase-shifted input remains in the argument.
Why the rules work
The justification uses the chain rule: the derivative of a composition is the derivative of its outer function evaluated at the inner expression, multiplied by the derivative of that inner expression. The constant-multiple rule allows the fixed coefficient \(A\) to remain outside the derivative.
For either model, the argument is \(u(x)=x-\phi\). Since \(\phi\) is constant, \(u'(x)=1-0=1\).
The derivative of \(\sin u\) with respect to \(u\) is \(\cos u\), and the derivative of \(\cos u\) with respect to \(u\) is \(-\sin u\).
Multiply the outer derivative by \(u'(x)=1\), then multiply by the constant \(A\). This gives the two stated formulas.
For example, applying those steps explicitly to the sine model gives $$ \frac{d}{dx}\left[A\sin(x-\phi)\right] =A\cos(x-\phi)\cdot\frac{d}{dx}(x-\phi) =A\cos(x-\phi)\cdot 1 =A\cos(x-\phi). $$ The cosine calculation differs only in the outer derivative: $$ \frac{d}{dx}\left[A\cos(x-\phi)\right] =-A\sin(x-\phi)\cdot\frac{d}{dx}(x-\phi) =-A\sin(x-\phi). $$ The factor \(1\) is present because the input has the form \(x-\phi\) with a constant phase, not because phase shifts can always be ignored.
Worked Example: Differentiating a shifted sine model
Worked Example: Differentiating a shifted sine model
Find the derivative of \(f(x)=4\sin(x-\pi/6)\).
The amplitude coefficient is \(4\), and the phase constant is \(\phi=\pi/6\). By the shifted sine rule, $$ f'(x)=4\cos(x-\pi/6). $$ To check the result at a convenient input, set \(x=\pi/6\). The original argument is \(\pi/6-\pi/6=0\), so the derivative formula gives $$ f'(\pi/6)=4\cos(0)=4(1)=4. $$ Directly applying the chain rule at that input gives \(4\cos(0)\cdot 1=4\) as well. The result retains the factor \(4\), and the derivative of the inner expression is \(1\), as required.
How amplitude and phase affect the result
The amplitude coefficient multiplies the derivative. If the coefficient is doubled while the phase and input stay fixed, every value of the derivative is doubled. If \(A=0\), the model is constant and its derivative is zero, which also agrees with both formulas.
The phase constant remains in the derivative’s argument. It determines where the sine or cosine input takes particular values; it is not itself differentiated as a variable. For the sine model, at \(x=\phi\) the argument equals zero, so $$ f'(\phi)=A\cos(0)=A. $$ For the cosine model, at \(x=\phi\), $$ g'(\phi)=-A\sin(0)=0. $$ These evaluations are useful checks on the location and value of the derivative at the phase point.
The two derivative rules also show why the sign must be handled by the function type. A shifted sine produces a cosine with no additional minus sign. A shifted cosine produces a negative sine. The phase shift does not remove or reverse that distinction.
Worked Example: A negative phase shift in a cosine model
Worked Example: A negative phase shift in a cosine model
Differentiate \(g(x)=5\cos(x+\pi/4)\).
Write the argument in the prescribed form: $$ x+\frac{\pi}{4}=x-\left(-\frac{\pi}{4}\right). $$ Thus the phase constant is \(\phi=-\pi/4\). Applying the shifted cosine rule, $$ g'(x)=-5\sin\left(x-\left(-\frac{\pi}{4}\right)\right) =-5\sin(x+\pi/4). $$ For a check, evaluate at \(x=-\pi/4\). The argument is \(-\pi/4+\pi/4=0\), and hence $$ g'(-\pi/4)=-5\sin(0)=0. $$ This agrees with the rule at \(x=\phi\), where the derivative of the shifted cosine model is zero. The negative sign in the derivative comes from differentiating cosine, not from the negative value of the phase constant.
Worked Example: Comparing equivalent shifted forms
Worked Example: Comparing equivalent shifted forms
Differentiate \(h(x)=3\sin(x+\pi/2)\), then compare the result with the equivalent expression \(3\cos x\).
First identify the phase as \(\phi=-\pi/2\), because \(x+\pi/2=x-(-\pi/2)\). The shifted sine rule gives $$ h'(x)=3\cos(x+\pi/2). $$ Using the trigonometric identity \(\cos(x+\pi/2)=-\sin x\), this becomes $$ h'(x)=-3\sin x. $$ The identity \(\sin(x+\pi/2)=\cos x\) also gives \(h(x)=3\cos x\). Differentiating this equivalent expression directly yields $$ \frac{d}{dx}(3\cos x)=-3\sin x, $$ which matches the result from the shifted sine rule. As a numerical check at \(x=0\), both expressions for the derivative give $$ 3\cos(\pi/2)=0 \qquad\text{and}\qquad -3\sin(0)=0. $$ The agreement confirms both the phase handling and the derivative sign.
Common mistake: losing the cosine minus sign
A frequent error is to treat the sine and cosine derivative rules as if they differed only by exchanging the names of the functions. For example, the following calculation is wrong: $$ \frac{d}{dx}\left[2\cos(x-\pi/3)\right] =2\sin(x-\pi/3). $$ It omits the negative sign from the derivative of cosine. The corrected calculation is $$ \frac{d}{dx}\left[2\cos(x-\pi/3)\right] =-2\sin(x-\pi/3)\cdot\frac{d}{dx}(x-\pi/3) =-2\sin(x-\pi/3)\cdot 1 =-2\sin(x-\pi/3). $$ A quick check at \(x=\pi/3\) distinguishes the results: the correct derivative is \(-2\sin(0)=0\), while the incorrect expression also happens to be zero there. Therefore, that particular input cannot detect the sign error. At \(x=\pi/3+\pi/2\), the correct expression is \(-2\sin(\pi/2)=-2\), whereas the incorrect one is \(2\). Choosing a check where the relevant sine is nonzero makes the sign visible.
A second risk is misreading the phase. For \(x+\pi/4\), the phase in the form \(x-\phi\) is \(-\pi/4\), not \(+\pi/4\). The safest procedure is to rewrite the argument as \(x-\phi\) before applying the rule, then preserve that exact argument in the derivative.
A reliable procedure
For a model with one shifted sine or cosine, the calculation can be organized as follows:
- Confirm that the model is a constant multiple of sine or cosine and that its phase is constant.
- Record the complete argument, preferably as \(x-\phi\).
- Retain the amplitude coefficient.
- Change sine to cosine, or cosine to negative sine.
- Keep the same shifted argument and check that its derivative is \(1\).
- Check the result at a convenient input, especially \(x=\phi\) or a point where the sine or cosine is \(1\) or \(-1\).
This procedure makes the separate roles visible: the outer function determines the sine-versus-cosine change and its sign, the amplitude multiplies the result, and the constant phase remains in the argument.
Why these derivatives matter
Derivative formulas describe how the output of a model changes as its input changes. Even before applying them in a physical or geometric setting, it is useful to understand what the formula preserves and what it changes. A constant amplitude does not disappear, and a constant horizontal shift does not vanish from the expression. Differentiation changes the trigonometric function according to its standard rule while keeping the shifted input.
This also gives a useful consistency check when working with equivalent trigonometric forms. If two expressions represent the same function, differentiating them by different valid routes should produce equivalent results. The third example illustrated that check. Such comparisons can reveal a missed phase sign or a missing negative sign.
Check Your Understanding
For each question, identify the amplitude and phase where appropriate, then apply the stated conditions carefully.
- Differentiate \(p(x)=6\sin(x-\pi/3)\). What is \(p'(\pi/3)\)?
- Differentiate \(q(x)=4\cos(x+\pi/6)\). Write its phase in the form \(x-\phi\).
- For \(r(x)=A\sin(x-\phi)\), with constant \(A\) and \(\phi\), find \(r'(\phi)\).
- A student claims that the derivative of \(2\cos(x-\pi/4)\) is \(2\sin(x-\pi/4)\). Identify the error and give the corrected derivative.
- Explain why the derivative of \(5\sin(x-\pi/2)\) retains the phase \(\pi/2\), even though the phase is constant.
- State two conditions required for the derivative rules in this tutorial to apply as written.
