Derivative of csc(x)
Cosecant is the reciprocal of sine, so its derivative comes straight from the derivative of sin(x) and one application of the chain rule.
Step-by-step solution
csc(x)=1/sin(x)=(sin x)^(−1). The outer function is u^(−1) with u=sin(x).
The outer derivative is −u^(−2); the inner derivative is cos(x). So f′(x)=−(sin x)^(−2)·cos(x)=−cos(x)/sin²(x).
−cos(x)/sin²(x)=−[1/sin(x)]·[cos(x)/sin(x)]=−csc(x)·cot(x).
Domain and graph behavior
csc(x) exists only where sin(x)≠0, so x≠kπ for every integer k; the derivative has the same domain, and each excluded point is a vertical asymptote. On (0, π) the function is positive: cot(x)>0 on (0, π/2) makes the derivative negative, and cot(x)<0 on (π/2, π) makes it positive. The branch therefore has a minimum at x=π/2, where csc=1 and the tangent is horizontal (y=1). On (π, 2π) the pattern flips and x=3π/2 is a local maximum with value −1.
Second derivative
Differentiate −csc(x)cot(x) with the product rule: f″(x)=csc(x)cot²(x)+csc³(x)=csc(x)[cot²(x)+csc²(x)]. Using cot²=csc²−1 this is csc(x)(2csc²(x)−1). The bracket is always positive, so the concavity follows the sign of csc(x): concave up where sin(x)>0, concave down where sin(x)<0. At x=π/2 the value is 1, confirming the minimum.
Common mistake
Do not take the reciprocal of the derivative of sine: d/dx[1/sin(x)] is not 1/cos(x). Differentiate the reciprocal with the chain or quotient rule. Also keep the minus sign; it is what separates csc′ from sec′=sec(x)tan(x).
Practice problems
Differentiate csc(3x)
Answer: −3csc(3x)cot(3x): the inner derivative 3 multiplies the outer result.
Differentiate 5csc(x)−x
Answer: −5csc(x)cot(x)−1: constant multiples stay, and the derivative of x is 1.
The co-function sign pattern
The three co-functions—cosine, cotangent and cosecant—all have derivatives that begin with a minus sign. Each mirrors its partner: sin′=cos becomes cos′=−sin, tan′=sec² becomes cot′=−csc², and sec′=sec·tan becomes csc′=−csc·cot. Replace every function by its co-function and attach a minus sign. Remembering this pattern halves the list of trigonometric derivatives you need to memorize.
A quotient-rule check
Treat csc(x) as 1/sin(x) and use the quotient rule: [0·sin(x)−1·cos(x)]/sin²(x)=−cos(x)/sin²(x). That is exactly the expression in step 2, so the chain-rule route and the quotient-rule route agree before any trigonometric rewriting.
A numerical check
At x=π/4, csc=√2 and cot=1, so the formula gives f′(π/4)=−√2≈−1.4142. A centered secant [csc(π/4+h)−csc(π/4−h)]/(2h) with h=0.0001 returns −1.41421. At x=π/6, csc=2 and cot=√3, giving −2√3≈−3.4641; the secant agrees to five decimal places. The slope steepens as x moves toward 0, matching the vertical asymptote.
More worked variations
csc(x²): the inner derivative is 2x, giving −2x·csc(x²)cot(x²).
csc²(x): square first, then chain: 2csc(x)·(−csc(x)cot(x))=−2csc²(x)cot(x).
x·csc(x): product rule gives csc(x)−x·csc(x)cot(x)=csc(x)(1−x·cot(x)).
Questions students ask
Is −csc(x)cot(x) the same as −cos(x)/sin²(x)?
Yes. They are the same function written two ways, valid wherever sin(x)≠0. Textbooks prefer −csc(x)cot(x); the sine-cosine form is convenient for evaluating signs.
Why is there no derivative at x=0?
csc(0) is undefined because sin(0)=0. A function cannot have a derivative at a point outside its domain, and the graph has a vertical asymptote there.
Related trigonometric derivatives
Review the trigonometric derivatives guide and the chain rule, or read our verification method and limitations.