Why Differentiate a Trigonometric Sum Term by Term?
A function may be built by adding or subtracting several trigonometric expressions. Its derivative records the combined rate of change, but the calculation does not require a special rule for each possible combination. The sum, difference, and constant-multiple rules let us differentiate one term at a time and then assemble the results.
This becomes especially useful when the terms have different inner functions. For instance, one term might involve \(\sin(2x)\), while another involves \(\cos(x^2)\). Each term has its own derivative, including its own chain-rule factor. Linearity then combines those derivatives with the original signs and constant coefficients.
Throughout, trigonometric derivative formulas use angles measured in radians. The sine and cosine functions are differentiable for every real input. Other trigonometric functions have restricted domains, so a sum containing one of them is defined only where all its terms are defined.
Linearity for Sums, Differences, and Constant Multiples
These rules apply to functions, not just to trigonometric functions. For trigonometric sums, the key practical step is to find the derivative of each term correctly before adding the results. If a term has the form \(\sin(u(x))\), \(\cos(u(x))\), or another composite trigonometric function, use its appropriate chain-rule formula.
When a summand is tangent, cotangent, secant, or cosecant, its input must also lie in that function's domain. The whole sum is differentiable on points where all its summands are defined and differentiable. In particular, \(\tan(u(x))\) and \(\sec(u(x))\) require \(\cos(u(x))\ne 0\); \(\cot(u(x))\) and \(\csc(u(x))\) require \(\sin(u(x))\ne 0\).
Why the Sum and Constant Rules Work
The rules follow from the definition of the derivative. For a function \(F\), its derivative at \(x\), when the limit exists, is \(\displaystyle F'(x)=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\). For two differentiable functions \(f\) and \(g\), apply this definition to their sum:
The last equality is valid because both derivative limits exist, and the limit of a sum is the sum of the limits. Replacing the plus sign between the functions with a minus sign gives the difference rule by the same argument.
For a constant multiple, take a constant \(c\) and factor it out of the difference quotient:
Repeated application proves the rule for any finite sum. The trigonometric formulas and the chain rule supply the derivatives of individual terms; linearity explains why those derivatives can then be added or subtracted.
Worked Examples
Worked Example: A Sum with Constant Coefficients
Differentiate \(f(x)=4\sin x-3\cos x+2\tan x\), and state its domain.
The sine and cosine terms are defined for every real \(x\). The tangent term requires \(\cos x\ne 0\), so \(f\) is defined and differentiable where \(x\ne \frac{\pi}{2}+k\pi\), for \(k\in\mathbb Z\).
Differentiate each term. The derivative of \(4\sin x\) is \(4\cos x\). The derivative of \(-3\cos x\) is \(-3(-\sin x)=3\sin x\). The derivative of \(2\tan x\) is \(2\sec^2 x\). Therefore, on the domain of \(f\),
Check each coefficient and sign against its term: the sine term contributes \(4\cos x\); the negative cosine term contributes \(3\sin x\); and the tangent term contributes \(2\sec^2 x\). These are precisely the terms in the result. The derivative has the same domain restriction as the original tangent term.
Worked Example: Different Inner Functions in the Summands
Differentiate \(g(x)=5\sin(2x-\frac{\pi}{3})+3\cos(x^2)-2\csc(4x)\), and give its domain.
The sine and cosine terms are defined for all real \(x\). The cosecant term requires \(\sin(4x)\ne 0\), so the full function and its derivative are defined where \(x\ne \frac{k\pi}{4}\), for \(k\in\mathbb Z\).
For the first term, the inner function \(2x-\frac{\pi}{3}\) has derivative \(2\). Thus the chain rule gives \(5\cos(2x-\frac{\pi}{3})\cdot 2=10\cos(2x-\frac{\pi}{3})\).
For the second term, the inner function \(x^2\) has derivative \(2x\). Therefore, its derivative is \(3[-\sin(x^2)](2x)=-6x\sin(x^2)\).
For the third term, the derivative formula is \(\frac{d}{dx}\csc(u(x))=-\csc(u(x))\cot(u(x))u'(x)\). Here \(u(x)=4x\), so \(u'(x)=4\). Including the coefficient \(-2\), the contribution is \((-2)[-\csc(4x)\cot(4x)](4)=8\csc(4x)\cot(4x)\). Combining all three contributions gives
As a check on the chain-rule factors, the inner derivatives are \(2\), \(2x\), and \(4\), respectively. The resulting coefficients and signs in the displayed derivative match those factors and the original coefficients. The result is valid wherever \(\sin(4x)\ne 0\).
Worked Example: Negative Coefficients and Affine Inputs
Differentiate \(h(x)=-\frac12\tan(3x+\frac{\pi}{4})+7\sec(\frac{x}{2})-4\cos(5-2x)\), and evaluate the derivative at \(x=0\).
The tangent term requires \(\cos(3x+\frac{\pi}{4})\ne 0\), and the secant term requires \(\cos(\frac{x}{2})\ne 0\). At \(x=0\), these conditions hold because the relevant cosines are \(\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}\) and \(\cos(0)=1\).
The inner derivative of \(3x+\frac{\pi}{4}\) is \(3\), so the first term contributes \(-\frac12\cdot 3\sec^2(3x+\frac{\pi}{4})=-\frac32\sec^2(3x+\frac{\pi}{4})\).
The inner derivative of \(\frac{x}{2}\) is \(\frac12\), so the second term contributes \(7\cdot\frac12\sec(\frac{x}{2})\tan(\frac{x}{2})=\frac72\sec(\frac{x}{2})\tan(\frac{x}{2})\).
For the last term, the inner derivative of \(5-2x\) is \(-2\). The derivative of \(\cos(5-2x)\) is \(-\sin(5-2x)(-2)=2\sin(5-2x)\); multiplying by \(-4\) gives \(-8\sin(5-2x)\). Thus
At \(x=0\), \(\sec^2(\frac{\pi}{4})=2\), \(\sec(0)=1\), \(\tan(0)=0\), and \(\sin(5-2\cdot 0)=\sin 5\). Therefore,
Checking the substitution term by term gives \(-3\) from the tangent term, \(0\) from the secant term, and \(-8\sin 5\) from the cosine term. Their sum is \(-3-8\sin 5\), as calculated.
A Common Mistake: Losing a Sign or an Inner Derivative
Linearity does not mean that the derivative of every summand is simply the derivative of its outer trigonometric function. Each composite term still needs its own chain-rule factor, and a negative coefficient stays attached to the derivative of that term.
For example, a mistaken calculation for \(q(x)=\sin(3x)-2\cos x\) might be \(q'(x)=\cos(3x)-2\sin x\). This omits the derivative \(3\) of the inner input \(3x\), and it gives the wrong sign for the derivative of \(-2\cos x\). Correcting both terms gives
The subtraction in the original expression applies to the whole second term. Since \(\frac{d}{dx}\cos x=-\sin x\), subtracting \(-2\cos x\)'s derivative contribution produces \(+2\sin x\). A reliable habit is to write the derivative of each complete summand, including its coefficient and inner derivative, before combining terms.
What to Keep in View
For a finite sum or difference of trigonometric functions, differentiate each summand on its own, retain its coefficient and sign, and then combine the results. When an argument depends on \(x\), multiply by the derivative of that argument. The sum rule applies only at points where the component functions are defined and differentiable; restrictions from tangent, cotangent, secant, and cosecant must not be discarded.
This procedure separates the work into two manageable tasks: finding each term's derivative and applying linearity. It also makes errors easier to locate. If the final answer seems doubtful, compare each derivative term with its original summand and check the coefficient, sign, and inner derivative separately.
Check Your Understanding
For each question, keep track of the signs, inner derivatives, and domains where they matter.
- Differentiate \(r(x)=3\sin x+5\cos x\).
- Differentiate \(s(x)=2\sin(4x)-\cos(x^2)\).
- Differentiate \(t(x)=-3\tan(2x)+\frac12\csc x\). Where is \(t\) defined?
- In differentiating \(-4\cos(3x+1)\), what are the contributions of the coefficient and the inner derivative?
- Explain why a sum containing \(\sec(2x)\) is not defined at every real \(x\), even if its other summands are defined everywhere.
