What Does a Power of a Trigonometric Function Mean?
A power of a trigonometric function combines two operations: first evaluate the trigonometric function, then raise its value to a power. For example, \(\sin^2(x)\) means the square of \(\sin(x)\), not the sine of \(x^2\). More explicitly,
This distinction matters when differentiating. The expression is a composition: the sine function is inside the square function. The power rule differentiates the outer power, and the Chain Rule accounts for the inner function.
The same structure appears when the angle is itself a function of \(x\). For instance, \(\tan^3(2x-1)\) means \((\tan(2x-1))^3\). In this case the calculation has three layers: a cube, a tangent, and the inner argument \(2x-1\). The Chain Rule carries the derivative through each layer.
General Rule for Positive Integer Powers
Let \(n\) be a positive integer, and let \(u\) be a differentiable function of \(x\). Wherever the trigonometric expression is defined, the following formulas combine the power rule, the standard trigonometric derivatives, and the Chain Rule.
| Function | Derivative | Where the formula applies |
|---|---|---|
| \(\sin^n(u)\) | \(n\sin^{n-1}(u)\cos(u)u'\) | Where \(u\) is defined |
| \(\cos^n(u)\) | \(-n\cos^{n-1}(u)\sin(u)u'\) | Where \(u\) is defined |
| \(\tan^n(u)\) | \(n\tan^{n-1}(u)\sec^2(u)u'\) | Where \(\cos(u)\ne 0\) |
| \(\cot^n(u)\) | \(-n\cot^{n-1}(u)\csc^2(u)u'\) | Where \(\sin(u)\ne 0\) |
| \(\sec^n(u)\) | \(n\sec^n(u)\tan(u)u'\) | Where \(\cos(u)\ne 0\) |
| \(\csc^n(u)\) | \(-n\csc^n(u)\cot(u)u'\) | Where \(\sin(u)\ne 0\) |
Here \(u'\) denotes the derivative of \(u\) with respect to \(x\). Sine and cosine are defined for every real input. Tangent and secant require a nonzero cosine in their input, while cotangent and cosecant require a nonzero sine. The derivative formula is only used at inputs where the original function is defined.
For \(n=1\), these rules reduce to the usual trigonometric derivative rules, with the factor \(u'\) included for a composite input. For \(n=0\), the function is the constant \(1\), so its derivative is \(0\); the displayed formulas are stated for \(n\geq 1\).
Why the Rule Works
Consider \(y=\sin^n(u(x))\). Define an outer function \(P(t)=t^n\) and an inner function \(v(x)=\sin(u(x))\). Then \(y=P(v(x))\). The ordinary power rule gives \(P'(t)=nt^{n-1}\), and the Chain Rule gives \(y'=P'(v(x))v'(x)\). Applying the Chain Rule again to \(v(x)=\sin(u(x))\) gives \(v'(x)=\cos(u(x))u'(x)\). Substitution yields the stated formula.
This argument explains the key structure of every formula in the table: differentiate the outer power, differentiate the trigonometric function, and multiply by the derivative of its input. For cosine, tangent, cotangent, secant, or cosecant, the same composition argument applies, using that function’s derivative and its domain conditions.
Worked Examples
Worked Example: Differentiating the Square of Sine
Differentiate \(y=\sin^2(x)\).
First identify the power \(n=2\) and the inner function \(u=x\), so \(u'=1\). The sine-power rule gives
To check the calculation, view the expression as the square of \(v(x)=\sin(x)\). The power rule gives \(2v(x)v'(x)\), and \(v'(x)=\cos(x)\). Therefore the derivative is \(2\sin(x)\cos(x)\), as obtained above. The input \(x\) can be any real number because sine is defined for every real number.
Worked Example: A Cubed Tangent with an Affine Input
Differentiate \(y=\tan^3(2x-1)\). The inner argument is \(u=2x-1\), so \(u'=2\). The function is defined only where \(\cos(2x-1)\ne 0\). With \(n=3\), apply the tangent-power rule:
Check the layers separately. The derivative of the outer cube \(t^3\), evaluated at \(t=\tan(2x-1)\), is \(3\tan^2(2x-1)\). The derivative of \(\tan(2x-1)\) is \(\sec^2(2x-1)\cdot 2\). Multiplying these factors gives the displayed derivative. The factor \(2\) comes from differentiating the inner argument.
Worked Example: A Cosine Power with a Quadratic Input
Differentiate \(y=\cos^4(3x^2)\). Here \(u=3x^2\), so \(u'=6x\). Use the cosine-power rule, including the minus sign in the derivative of cosine:
For a check, set \(v(x)=\cos(3x^2)\), so \(y=v(x)^4\). Differentiating the outer power gives \(4v(x)^3\). Differentiating \(v\) gives \(-\sin(3x^2)\cdot 6x\). Their product is \(-24x\cos^3(3x^2)\sin(3x^2)\), exactly as above. This function is defined for every real \(x\).
Worked Example: A Squared Cotangent
Differentiate \(y=\cot^2(5x)\). The inner function is \(u=5x\), with \(u'=5\). Cotangent is defined here when \(\sin(5x)\ne 0\). The cotangent-power rule gives
To check, differentiating the outer square contributes \(2\cot(5x)\). Differentiating \(\cot(5x)\) contributes \(-\csc^2(5x)\cdot 5\). Multiplying gives \(-10\cot(5x)\csc^2(5x)\), with the negative sign and inner factor both present.
A Reliable Differentiation Procedure
When differentiating a power of a trigonometric function, identify the layers before writing the derivative. This prevents the power, trigonometric derivative, and inner derivative from being confused.
For \(\sin^5(4x)\), the power is \(5\), and the base is \(\sin(4x)\).
Bring the exponent down and reduce it by one: \(5\sin^4(4x)\).
For sine this supplies \(\cos(4x)\); for tangent it supplies \(\sec^2(4x)\), and so on.
Multiply by the derivative of \(4x\), which is \(4\). For a more complicated input, find its derivative before completing the product.
Ensure the original trigonometric function is defined at the input, and check for a negative sign in derivatives such as those of cosine and cotangent.
Common Mistake: Omitting the Inner Derivative
A frequent error is to apply the power rule and differentiate the trigonometric function, but forget to differentiate its input. For example, the following calculation is incorrect:
The expression inside the sine is \(4x\), whose derivative is \(4\), not \(1\). The correction is
Checking by layers confirms the correction: differentiating the outer cube contributes \(3\sin^2(4x)\), while differentiating \(\sin(4x)\) contributes \(\cos(4x)\cdot 4\). Their product has the factor \(12\), so the incorrect result is missing a factor of \(4\).
A related notation error is to read \(\sin^2(x)\) as \(\sin(x^2)\). These are different functions. The first squares a sine value; the second takes the sine of a squared input. Make the intended composition clear before differentiating.
Why These Derivatives Matter
Powers of trigonometric functions arise whenever a periodic quantity is transformed by a power, or when a trigonometric expression is nested inside another function. Their derivatives describe how the powered output changes as the input changes. The calculation is not a new derivative rule for each particular exponent and angle: it is a consistent application of the power rule and Chain Rule.
The main practical lesson is to preserve every layer. A correct derivative includes the outer exponent, the derivative of the trigonometric function, and the derivative of its argument. It also respects the domain of the original function. Keeping these pieces visible makes both the calculation and its verification more reliable.
Check Your Understanding
For each question, identify the power and inner argument before differentiating. Include any domain restriction that applies.
- Differentiate \(y=\sin^4(x)\).
- Differentiate \(y=\cos^3(2x+1)\).
- Differentiate \(y=\tan^2(3x)\), and state where the original function is defined.
- Differentiate \(y=\sec^3(x^2)\), and state its domain condition.
- What factor is missing from the incorrect derivative \(4\cos^3(5x)\sin(5x)\) of \(\cos^4(5x)\)? Write the corrected derivative.
