Why Order Matters
A trigonometric expression can contain several layers of differentiation at once. For example, a sine function may have a polynomial input, its value may be raised to a power, and the result may be multiplied by another trigonometric function. The rules are familiar individually, but applying them in an unclear order makes it easy to omit a factor or differentiate only part of the expression.
A reliable approach is to begin with the outermost operation. If the expression is a product, use the product rule first; if it is a quotient, use the quotient rule first; if it is a power, use the power rule first. Then differentiate each resulting factor through its inner functions using the chain rule.
The Rules Used in Mixed Problems
We use the standard trigonometric derivatives, with angles measured in radians. If \(u=u(x)\) is differentiable, then
These are chain-rule formulas: the derivative of the outer trigonometric function is multiplied by the derivative of its input. The tangent formula applies wherever \(\tan u\) is defined, which requires \(\cos u\ne 0\).
For differentiable functions \(a\) and \(b\), the product rule is
For a quotient, the quotient rule is
For a positive integer \(n\), the power rule combined with the chain rule gives
In mixed problems, these rules are not competing alternatives. A product can have a factor that is itself a power, and that power can contain a nested trigonometric function. Apply the rule for the outer operation, then continue inward.
Why the Chain Rule Contributes an Inner Derivative
The chain rule states that if \(g\) is differentiable at \(x\), and \(f\) is differentiable at \(g(x)\), then the composite function \(f(g(x))\) is differentiable at \(x\) and
Here is a justification using the definition of differentiability. Set \(\Delta=g(x+h)-g(x)\). Differentiability of \(f\) at \(g(x)\) means that the change in \(f\) can be written as
where \(r(\Delta)/\Delta\to 0\) as \(\Delta\to 0\), for nonzero \(\Delta\). Differentiability of \(g\) implies both that \(\Delta\to 0\) as \(h\to 0\) and that \(\Delta/h\to g'(x)\). Therefore, when \(\Delta\ne 0\), dividing the change in \(f(g(x))\) by \(h\) gives
The first term tends to \(f'(g(x))g'(x)\). In the second term, \(r(\Delta)/\Delta\to 0\), while \(\Delta/h\) has a finite limit, so the second term tends to zero. If \(\Delta=0\) for some \(h\), the change in \(f(g(x))\) is also zero for that \(h\), consistent with the same limiting result. Thus the derivative is \(f'(g(x))g'(x)\). The inner derivative is required because a change in \(x\) first changes \(g(x)\), and that change then affects \(f\).
A Reliable Working Order
Before differentiating, rewrite the expression mentally as a sequence of operations. For a product of a power of a nested trigonometric function and another factor, for example, the order is product, power, trigonometric function, inner function. Making that structure explicit prevents a common error: stopping after differentiating the visible trigonometric function while forgetting its input is also changing.
Decide whether the expression begins as a product, quotient, or power, or is a single composite function.
Keep all factors that the product, quotient, or power rule requires.
Move inward one layer at a time, multiplying by the derivative of each inner function.
Confirm that every product has both terms, every quotient has its denominator squared, and each nested input contributes its derivative.
Worked Examples
Worked Example: A Trigonometric Power Multiplied by Another Factor
Differentiate \(y=(\sin(2x+1))^3\cos x\). The outer operation is a product. Let \(A=(\sin(2x+1))^3\) and \(B=\cos x\), so \(y'=A'B+AB'\).
For \(A\), apply the power rule, then differentiate the sine, then its input:
For the other factor, \(B'=-\sin x\). Therefore, the product rule gives
Check at \(x=0\). The expression for the derivative gives \(6\sin^2(1)\cos(1)\cdot 1-\sin^3(1)\cdot 0=6\sin^2(1)\cos(1)\). Directly from the product-rule pieces, \(A'(0)B(0)+A(0)B'(0)=6\sin^2(1)\cos(1)\cdot1+\sin^3(1)\cdot0\), the same value.
Worked Example: A Quotient with a Nested Tangent
Differentiate
Set \(N=\tan(\sin(2x))\) and \(D=1+x^2\). The quotient rule requires \(D\ne 0\); here \(1+x^2>0\) for every real \(x\). First differentiate the numerator, moving from the outer tangent to its sine input and then to \(2x\):
Also, \(D'=2x\). Substituting both derivatives into the quotient rule gives
Check at \(x=0\). The numerator of the original quotient is \(\tan(\sin 0)=0\), its derivative is \(2\cos 0\sec^2(\sin 0)=2\), and the denominator is \(1\) with derivative \(0\). The quotient rule therefore gives \((2\cdot1-0\cdot0)/1^2=2\). The displayed derivative also evaluates to \(2\) at \(x=0\).
Worked Example: A Power of a Cosine with a Linear Input
Differentiate \(y=(\cos(1-4x))^4\). The outer operation is a fourth power. Let \(v=1-4x\); then \(v'=-4\). Apply the power rule and then the chain rule:
Check at \(x=0\). The displayed derivative is \(16\cos^3(1)\sin(1)\). Independently, if \(w=\cos(1-4x)\), then \(w'=-\sin(1-4x)(-4)=4\sin(1-4x)\), so differentiating \(w^4\) gives \(4w^3w'=16\cos^3(1-4x)\sin(1-4x)\) at \(x=0\), as required.
Worked Example: A Product with Two Nested Trigonometric Functions
Differentiate \(y=\sin(\cos(3x))\tan(2x)\). Apply the product rule. For the first factor, differentiate the outer sine, then the cosine, then \(3x\):
For the second factor, the tangent chain rule gives \(\frac{d}{dx}\tan(2x)=2\sec^2(2x)\). Thus
Check at \(x=0\). The first product-rule term is zero because \(\sin(0)=0\) and \(\tan(0)=0\). The second is \(2\sin(\cos 0)\sec^2 0=2\sin(1)\). Directly, the first factor has value \(\sin(1)\) and the second factor has value \(0\); their derivatives at zero are \(0\) and \(2\), respectively. The product rule therefore gives \(0\cdot0+\sin(1)\cdot2=2\sin(1)\).
A Common Mistake: Losing an Inner Factor
Consider again \(N=\tan(\sin(2x))\). A common incorrect calculation is
This differentiates the outer tangent but stops there. It omits the derivative of the input \(\sin(2x)\). Continuing inward gives \(\frac{d}{dx}\sin(2x)=2\cos(2x)\), so the correct derivative is
The omission changes the value: at \(x=0\), the incorrect expression gives \(1\), whereas the correct derivative gives \(2\). When the chain rule is combined with a product or quotient rule, this missing factor remains missing unless each factor is checked separately.
Putting the Method to Work
Mixed trigonometric derivatives appear whenever a changing quantity is described through several linked operations. A reliable order makes the algebra easier to audit: name the factors, identify the outer rule, and then work inward through each nested function. Simplifying the final expression is optional; preserving every factor correctly is essential.
Before accepting an answer, ask whether it contains the required terms from the outer rule and the derivative of every inner input. For quotients, also check that the denominator is nonzero on the domain being considered. These checks are especially useful when several trigonometric functions appear in the same expression.
Check Your Understanding
For each question, identify the first rule you would apply before carrying out the differentiation.
- Differentiate \(y=\sin^2(5x)\cos x\). Which rule applies to the whole expression first?
- Differentiate \(y=\frac{\cos(x^2)}{2+x}\). What condition must hold for the quotient rule?
- Differentiate \(y=\tan^3(1-2x)\). List the layers of differentiation from outermost to innermost.
- In differentiating \(\sin(\cos(4x))\), which inner derivatives must be included?
- For a proposed derivative of a mixed expression, name two structural checks that can reveal a missing term or factor.
