Why Products of Trigonometric Powers Need Two Derivative Steps
Expressions such as \(\sin^2 x\cos^3 x\) contain two kinds of structure at once. Each factor is a power of a trigonometric function, and the factors are multiplied together. Differentiating the powers calls for the power rule and Chain Rule; differentiating their product calls for the product rule.
The order of these steps matters. First find the derivative of each factor, including the derivative of its inner function. Then combine those derivatives using the product rule. This method also works when the two factors have different inputs, such as \(\sin^2(2x+1)\cos^3(x^2)\).
The Rule for a Product of Sine and Cosine Powers
Let \(u\) and \(v\) be differentiable functions of \(x\), and let \(m\) and \(n\) be positive integers. Assume angles are measured in radians, so the standard trigonometric derivative formulas apply. The power rule and Chain Rule give the derivatives of the individual factors, and the product rule combines them:
Here \(u'\) and \(v'\) mean the derivatives of \(u(x)\) and \(v(x)\) with respect to \(x\). The formula keeps the two factors’ inputs separate: the first term differentiates the sine power and retains the cosine power, while the second differentiates the cosine power and retains the sine power.
For the common case where both powers have the same input, set \(u=v=x\). Then \(u'=v'=1\), and the formula becomes:
The exponents \(m\) and \(n\) in this rule are positive integers. This keeps the powers and their derivatives defined by repeated multiplication and the usual power rule, including when sine or cosine is zero.
Why the Formula Works
The justification follows directly from the product rule and the derivatives of the two factors. Write \(f(x)=\sin^m(u(x))\) and \(g(x)=\cos^n(v(x))\). The product rule states that \((fg)'=f'g+fg'\). By the power rule and Chain Rule, differentiating the first factor gives \(f'=m\sin^{m-1}(u)\cos(u)u'\). Differentiating the second gives \(g'=-n\cos^{n-1}(v)\sin(v)v'\), because the derivative of cosine is negative sine.
Each term has a clear source: the first is the derivative of the sine power times the unchanged cosine power; the second is the unchanged sine power times the derivative of the cosine power. The negative sign in the second term comes from differentiating cosine.
Worked Example: Differentiate \(\sin^2 x\cos^3 x\)
Worked Example: Differentiate \(\sin^2 x\cos^3 x\)
Let \(y=\sin^2 x\cos^3 x\). Apply the product rule, differentiating each factor in turn. The derivative of \(\sin^2 x\) is \(2\sin x\cos x\), and the derivative of \(\cos^3 x\) is \(-3\cos^2 x\sin x\).
Check the result at \(x=\pi/6\), where \(\sin x=1/2\) and \(\cos x=\sqrt{3}/2\). The displayed derivative gives:
Independently, the first product-rule term at this angle is \((2\sin x\cos x)\cos^3x=(\sqrt3/2)(3\sqrt3/8)=9/16\). The second is \(\sin^2x(-3\cos^2x\sin x)=(1/4)(-9/8)=-9/32\). Their sum is \(9/16-9/32=9/32\), matching the derivative formula.
When the Inputs Are Different
If the powers contain different functions of \(x\), do not treat their inputs as though they were the same. Each input has its own derivative, and each derivative belongs in the term where that input was differentiated. For example, the derivative of \(\sin^4(3x)\) contains a factor of \(3\), while the derivative of \(\cos^2(2x)\) contains a factor of \(2\).
Worked Example: Different Powers and Affine Inputs
Worked Example: Different Powers and Affine Inputs
Differentiate \(y=\sin^4(3x)\cos^2(2x)\). The derivative of the first factor is \(4\sin^3(3x)\cos(3x)\cdot 3\). The derivative of the second factor is \(2\cos(2x)(-\sin(2x))\cdot 2\). The product rule therefore gives:
Check at \(x=\pi/6\). Then \(3x=\pi/2\) and \(2x=\pi/3\). In the first term, \(\cos(3x)=0\), so the term is zero. In the second term, \(\sin^4(3x)=1\), \(\cos(2x)=1/2\), and \(\sin(2x)=\sqrt3/2\). Thus:
For an independent product-rule check, at this point the derivative of the first factor is \(12\sin^3(3x)\cos(3x)=0\), and the second factor has value \(\cos^2(\pi/3)=1/4\). The first product-rule term is therefore \(0\cdot(1/4)=0\). The first factor has value \(1\), and the derivative of the second factor is \(-4\cos(\pi/3)\sin(\pi/3)=-\sqrt3\). The second product-rule term is \(1\cdot(-\sqrt3)=-\sqrt3\), confirming the answer.
Worked Example: Different Composite Inputs
Worked Example: Different Composite Inputs
Differentiate \(y=\sin^2(2x+\pi/6)\cos^3(x+\pi/3)\). Let \(u=2x+\pi/6\) and \(v=x+\pi/3\), so \(u'=2\) and \(v'=1\). Applying the product rule and Chain Rule gives:
Check at \(x=0\). The two inputs are \(\pi/6\) and \(\pi/3\). Substituting their exact trigonometric values into the result gives:
Check by differentiating the factors separately at \(x=0\). The sine-power factor has value \(1/4\) and derivative \(4(1/2)(\sqrt3/2)=\sqrt3\). The cosine-power factor has value \(1/8\) and derivative \(-3(1/2)^2(\sqrt3/2)=-3\sqrt3/8\). The product-rule terms are \((\sqrt3)(1/8)=\sqrt3/8\) and \((1/4)(-3\sqrt3/8)=-3\sqrt3/32\). Their sum is \(\sqrt3/32\), as obtained above.
A Common Mistake: Multiplying the Factor Derivatives
The derivative of a product is not the product of the derivatives. For example, a mistaken calculation for \(y=\sin^2x\cos^3x\) might be:
This expression differentiates both factors at once, but it omits the product rule’s two terms, each of which retains one undifferentiated factor. The correction is:
The incorrect expression also fails a simple check at \(x=0\): it gives zero, while the correct derivative formula gives zero there too, so that point does not distinguish them. At \(x=\pi/6\), however, the incorrect expression equals \((\sqrt3/2)(-9/8)=-9\sqrt3/16\), whereas the correctly computed derivative is \(9/32\). A check at a useful point can expose an error that happens to be hidden at a special input.
A Reliable Procedure
For a product of two trigonometric powers, use the same sequence each time. Keep the factor that is not being differentiated unchanged in each product-rule term.
Separate the expression into its sine-power factor and cosine-power factor.
Use the power rule and Chain Rule, including the derivative of each input.
Write the derivative of the first factor times the unchanged second factor, then add the unchanged first factor times the derivative of the second.
The derivative of cosine contributes a minus sign; composite inputs contribute their own derivatives.
These products occur whenever a function combines oscillating factors with different powers. The essential skill is not to memorize one expanded expression, but to keep track of which factor is differentiated in each term and to include every inner derivative. That discipline makes the same method dependable for equal or different powers and for equal or different inputs.
Check Your Understanding
For each question, identify the product-rule terms before simplifying.
- Differentiate \(\sin^3x\cos^2x\).
- Differentiate \(\sin^2(4x)\cos^3x\).
- In differentiating \(\sin^5(u(x))\cos^2(v(x))\), which term contains \(u'\), and which contains \(v'\)?
- Why does differentiating the cosine-power factor contribute a negative sign?
- Explain why the product of the two factor derivatives is not the derivative of their product.
