Why Products Need a Separate Rule
A trigonometric expression may contain two or more functions multiplied together. Differentiating such an expression requires accounting for how each factor changes. The product rule does this: for two differentiable factors, differentiate the first while leaving the second unchanged, then add the first unchanged times the derivative of the second.
When a factor also has a composite input, such as \(\sin(3x)\), its derivative must include the chain rule. Thus a product can require both rules at once. The main discipline is to identify the factors clearly before differentiating.
Trigonometric Derivatives Used in Products
The product rule does not change the derivative of an individual trigonometric factor. The usual trigonometric derivative formulas are combined with the chain rule when the input is not simply \(x\). For a differentiable function \(g\), the formulas we will use include:
The sine and cosine formulas apply wherever \(g\) is differentiable. The tangent formula requires \(\cos(g(x))\ne 0\), and the secant formula requires the same condition, so that the tangent or secant is defined. These conditions also apply when such factors occur in a product.
For a pair \(u(x)v(x)\), the product rule gives two terms, not one. If \(u(x)=\sin(g(x))\) and \(v(x)=\cos(h(x))\), then the inputs \(g\) and \(h\) need not be the same:
The first term comes from differentiating the sine factor while keeping the cosine factor in place. The second comes from differentiating the cosine factor while keeping the sine factor in place. In particular, \(g'(x)\) and \(h'(x)\) belong to different terms.
Worked Example: A Product of Sine and Cosine
Differentiate \(y=\sin x\cos x\).
Take \(u(x)=\sin x\) and \(v(x)=\cos x\). Their derivatives are \(u'(x)=\cos x\) and \(v'(x)=-\sin x\). Apply the product rule:
To check the calculation at \(x=0\), the derivative formula gives \(\cos^2(0)-\sin^2(0)=1^2-0^2=1\). The two product-rule contributions at \(0\) are \(\cos(0)\cos(0)=1\) and \(\sin(0)(-\sin(0))=0\), which also sum to \(1\). No trigonometric identity is needed to state the derivative.
Products with Different Composite Inputs
When the factors have different inputs, work out each derivative separately. A common source of errors is to use the derivative of the outer trigonometric function but omit the derivative of its input. For example, differentiating \(\sin(3x)\) produces the factor \(3\), while differentiating \(\cos(x^2+1)\) produces the factor \(2x\).
Worked Example: Sine and Cosine with Different Inputs
Differentiate \(y=\sin(3x)\cos(x^2+1)\).
Let \(u(x)=\sin(3x)\) and \(v(x)=\cos(x^2+1)\). By the chain rule,
Now apply the product rule and substitute both factor derivatives:
Check at \(x=0\). The first term is \(3\cos(0)\cos(1)=3\cos(1)\). The second term is \(-2(0)\sin(0)\sin(1)=0\). Thus the formula gives \(y'(0)=3\cos(1)\). Directly from the product rule at \(0\), \(u'(0)v(0)+u(0)v'(0)=3\cdot\cos(1)+0\cdot v'(0)=3\cos(1)\), confirming the evaluation.
Why the Product Rule Works
The rule follows from the definition of derivative. Suppose \(u\) and \(v\) are differentiable at \(x\). In the difference quotient for their product, add and subtract \(u(x+h)v(x)\). For nonzero \(h\), this gives:
As \(h\) approaches \(0\), differentiability of \(u\) implies continuity of \(u\) at \(x\), so \(u(x+h)\) approaches \(u(x)\). The difference quotients for \(v\) and \(u\) approach \(v'(x)\) and \(u'(x)\), respectively. Therefore the limit of the displayed expression is \(u(x)v'(x)+v(x)u'(x)\). This is the derivative of \(uv\), which proves the product rule. The argument applies to trigonometric factors because, wherever their inputs are defined and differentiable, their compositions are differentiable as well.
Products of Three or More Factors
For three differentiable factors \(u\), \(v\), and \(w\), apply the two-factor rule and then apply it again to the resulting products. This gives:
Each term differentiates exactly one factor and leaves every other factor unchanged. For any positive integer \(n\), the same pattern holds for a product of \(n\) differentiable factors: the derivative is the sum of \(n\) terms, with one factor differentiated in each term. This follows by induction: it holds for two factors by the product rule; if it holds for \(n\) factors, apply the product rule to their product times one more factor. The result consists of the existing \(n\) terms multiplied by the new factor, together with one term in which the new factor is differentiated.
Worked Example: A Product of Three Trigonometric Factors
Differentiate \(y=\sin(2x)\cos x\sec(3x)\). This function is defined where \(\cos(3x)\ne 0\), because the secant factor must be defined.
Name the factors \(u(x)=\sin(2x)\), \(v(x)=\cos x\), and \(w(x)=\sec(3x)\). Their derivatives are:
Differentiate one factor at a time, retaining the other two in each term:
Check at \(x=0\), where all factors and derivatives are defined. The first term is \(2\cos(0)\cos(0)\sec(0)=2\). The second term is \(-\sin(0)\sin(0)\sec(0)=0\). The third term is \(3\sin(0)\cos(0)\sec(0)\tan(0)=0\). Thus \(y'(0)=2\). Substituting \(x=0\) into the three-factor rule before expansion gives \(u'(0)v(0)w(0)+u(0)v'(0)w(0)+u(0)v(0)w'(0)=2\cdot1\cdot1+0+0=2\), confirming the result.
A Common Product-Rule Mistake
Consider \(y=\sin(3x)\cos x\). A tempting but incorrect calculation differentiates both factors in the same term:
This expression omits the two-factor product rule’s second contribution. It behaves as though the change in the product came only from changing both factors simultaneously. The product rule instead accounts for the change in each factor separately, while holding the other factor in place in that term. The correct calculation is:
At \(x=0\), the incorrect expression gives \(3\cdot1\cdot0=0\), whereas the correct expression gives \(3\cdot1\cdot1-0\cdot0=3\). The discrepancy makes the omitted term visible. A reliable check is to count factors: for two factors there must be two product-rule contributions; for three factors there must be three.
A Reliable Differentiation Process
Before simplifying, keep the product-rule structure visible. This makes it easier to check that every factor has been differentiated once and that the chain rule has been applied to each input.
Write the expression as \(f_1(x)f_2(x)\cdots f_n(x)\), and note any domain restrictions such as a secant or tangent denominator being nonzero.
Use the trigonometric derivative formula and include the derivative of every composite input.
In each term, differentiate one factor and leave all the others unchanged.
Count the terms, inspect chain-rule factors and signs, then test the formula at a convenient point in its domain.
Check Your Understanding
For each question, keep the product-rule contributions separate until every factor derivative has been found.
- Differentiate \(y=\cos x\sin(2x)\).
- Differentiate \(y=\sin(x^2)\cos(4x)\), showing both chain-rule factors.
- Differentiate \(y=\tan x\sec(2x)\) and state the domain restriction.
- How many terms occur when differentiating a product of five differentiable factors?
- For \(y=\sin x\cos x\sec x\), write the three product-rule terms without combining them.
