Derivative of cos²(x)
cos²(x) means (cos x)². Differentiate the square, then multiply by the derivative of cosine—which carries the minus sign into the answer.
Step-by-step solution
cos²(x)=(cos x)². Let u=cos(x); the outer function is u². In the calculator, type cos(x)^2.
The outer derivative is 2u; the inner derivative is −sin(x).
f′(x)=2cos(x)·(−sin(x))=−2sin(x)cos(x)=−sin(2x).
Domain and graph behavior
The function and its derivative are defined for every real x. Since cos²(x)=(1+cos 2x)/2, the graph has period π and values in [0, 1]. Maxima with value 1 occur at x=kπ and minima with value 0 at x=π/2+kπ. The function decreases on (0, π/2), where −sin(2x)<0, and increases on (π/2, π).
Second derivative
f″(x)=d/dx[−sin(2x)]=−2cos(2x). At x=0 it equals −2, confirming the maximum is concave down. It vanishes at x=π/4+kπ/2, where the function equals 1/2; those are the inflection points.
Common mistake
The most common slip is losing the minus sign from (cos x)′=−sin(x), which produces +2sin(x)cos(x). A quick check: cos²(x) falls from 1 toward 0 just after x=0, so the derivative must be negative there. Also do not confuse cos²(x) with cos(x²).
Practice problems
Differentiate cos²(5x)
Answer: 2cos(5x)·(−sin(5x))·5=−5sin(10x).
Differentiate 1−cos²(x)
Answer: sin(2x). This function equals sin²(x), so the result matches that page.
The same curve, shifted
cos(x)=sin(x+π/2), so cos²(x)=sin²(x+π/2): the graph of cos² is the graph of sin² moved left by π/2. Differentiating the shifted form gives sin(2x+π)=−sin(2x), the same answer. Shifting a graph shifts its slope function by the same amount.
Check with the power-reduction identity
Rewrite cos²(x)=(1+cos 2x)/2. The constant contributes nothing, and cos(2x)/2 differentiates to −(2sin 2x)/2=−sin(2x). This route avoids the outer power entirely.
Where cos² appears in applications
In alternating-current circuits, instantaneous power in a resistor is proportional to cos²(ωt). Its rate of change is d/dt cos²(ωt)=−ω·sin(2ωt): the chain rule contributes ω, and the power oscillates at twice the voltage frequency. Its average over a full period is 1/2, which is why root-mean-square values involve √2.
A numerical check
At x=π/4 the formula gives −sin(π/2)=−1. At x=π/3 it gives −sin(2π/3)=−√3/2≈−0.8660. Centered secants with h=0.0001 match both values to at least six decimal places.
More worked variations
x·cos²(x): cos²(x)−x·sin(2x).
cos³(x): −3cos²(x)sin(x).
cos²(x)−sin²(x): this equals cos(2x), so its derivative is −2sin(2x).
Questions students ask
Why do the derivatives of sin²(x) and cos²(x) add to zero?
Because sin²(x)+cos²(x)=1 is constant, and a constant has derivative zero. The two answers, sin(2x) and −sin(2x), cancel.
Do I need the double-angle identity?
No. −2sin(x)cos(x) is a complete, correct answer. Rewriting as −sin(2x) is optional simplification.
Related derivatives
Review the chain rule and the trigonometric derivatives guide, or read our verification method and limitations.