Derivative of acos(x)
Answer
Derivative of acos(x)
- Rule Used
- Inverse Trigonometric Rule
- Steps
- Recognize an inverse trigonometric function. It is not the reciprocal of the trig function: means , the angle whose sine is x.
- The derivative comes from implicit differentiation of the inverse relation, then a Pythagorean identity that removes the angle.
- Final answer: -
d/dx arccos(x) = d/dx cos^-1(x) = -1 / sqrt(1 - x^2), for -1 < x < 1.
Instant step-by-step solution for acos(x)
Derivative of acos(x) = -sqrt(-x^2+1)^(-1) Solve step-by-step for free: https://derivativecalculatorai.com/derivative-of-arccos-x
Formula for arccos(x) and cos^-1(x)
d/dx arccos(x) = d/dx cos^-1(x) = -1 / sqrt(1 - x^2), for -1 < x < 1.
The only difference from arcsin is the sign. That is not a convention: arccos decreases as x increases, so the slope has to be negative. The calculator is filled with acos(x).
Why the derivative of arccos is negative
Let y = arccos(x). Then y is the unique angle in [0, π] with cos(y) = x. Differentiating both sides, the chain rule produces -sin(y) · y' = 1.
Solving for y' puts the minus sign in the numerator: y' = -1/sin(y). On [0, π] the sine is non-negative, so sin(y) = sqrt(1 - cos^2(y)) = sqrt(1 - x^2). The negative sign stays in front of the root.
There is a second proof that does not repeat that algebra. The identity arcsin(x) + arccos(x) = π/2 holds for every x in [-1, 1]. The derivative of a constant is 0, so the two derivatives add to 0. Once the derivative of arcsin is 1/sqrt(1 - x^2), the derivative of arccos is the negative of it.
- Set y = arccos(x), so cos(y) = x and y is in [0, π].
- Differentiate both sides: -sin(y) · dy/dx = 1.
- Solve: dy/dx = -1/sin(y).
- On [0, π], sin(y) ≥ 0, so sin(y) = sqrt(1 - cos^2(y)) = sqrt(1 - x^2).
- Conclude d/dx arccos(x) = -1/sqrt(1 - x^2), for -1 < x < 1.
- Check: add the derivative of arcsin(x). The sum is 0, matching arcsin(x) + arccos(x) = π/2.
Worked examples that keep the minus sign
Slope of arccos at the origin
- Substitute x = 0 into -1/sqrt(1 - x^2).
- The root is 1, so the slope is -1.
- Arccos crosses π/2 at the origin and is decreasing, which matches a slope of -1.
Answer: -1
Derivative of arccos(3x)
- Outer function: -1/sqrt(1 - u^2) with u = 3x.
- Inner derivative: 3. Multiply, and square the 3 inside the root.
- 1 - 9x^2 is the radicand. The domain is |3x| < 1, so |x| < 1/3.
Answer: -3 / sqrt(1 - 9x^2), for |x| < 1/3
Derivative of x · arccos(x)
- Product rule with u = x and v = arccos(x).
- u' = 1 and v' = -1/sqrt(1 - x^2).
- The second term is negative. Do not cancel it against the first.
Answer: arccos(x) - x/sqrt(1 - x^2)
Mistakes that erase the minus sign
Copying the arcsin formula unchanged
Arcsin and arccos have opposite slopes. An answer of 1/sqrt(1 - x^2) for arccos fails the identity check: the derivatives would add to 2/sqrt(1 - x^2) instead of 0.
Treating cos^-1(x) as sec(x)
cos^-1(x) is arccos(x). The reciprocal of cosine is sec(x), whose derivative is sec(x) tan(x). Secant is undefined at π/2, where arccos is perfectly smooth.
Losing the 3 in arccos(3x)
The chain rule multiplies by 3 and replaces x^2 with (3x)^2 = 9x^2. Writing -1/sqrt(1 - 9x^2) keeps the inside correct and still drops the factor in front.
Using the formula outside (-1, 1)
arccos(2) is not a real number, and neither is -1/sqrt(1 - 4). The expression under the root has to be positive, so the input of arccos, including any inside function, has to lie strictly between -1 and 1.
Related functions
- Derivative of arcsin(x) — The same root with the opposite sign. Their sum is the derivative of π/2, which is 0.
- Derivative of arctan(x) — 1/(1 + x^2). No square root, and no restriction to (-1, 1).
- Derivative of cos(x) — -sin(x). That minus sign is the ordinary cosine rule, not this inverse rule.
- Derivatives of trigonometric functions — Sin through cot, then the three inverse functions, in one table.
Frequently asked questions
What is the derivative of arccos(x)?
The derivative of arccos(x) is -1/sqrt(1 - x^2), for -1 < x < 1.
What is the derivative of cos^-1(x)?
cos^-1(x) means arccos(x). Its derivative is -1/sqrt(1 - x^2), not the derivative of sec(x).
Why is the derivative of arccos negative?
arccos decreases from π to 0 as x goes from -1 to 1, so every slope is negative. The identity arcsin(x) + arccos(x) = π/2 forces the two derivatives to be opposites.
How is the derivative of arccos related to arcsin?
They add to zero. d/dx arccos(x) = - d/dx arcsin(x) = -1/sqrt(1 - x^2).