01 · An exact arcsine gradient
Find the gradient of y = arcsin x at x = 1/2.
Hint
1 − (1/2)² = 3/4.
Worked solution
The gradient is 2/√3. The function value is π/6 radians.
Understand · explore · practise
Differentiating inverse trigonometric functions
Derive the derivatives of arcsin, arccos and arctan with correct branches and domains. Practise composite functions, products, quotients and tangents, with an extension on reciprocal inverse functions.
Before you startPrincipal inverse trig branches; inverse-function derivative; chain, product and quotient rules
01 / Choose the principal inverse branch
An inverse trig function returns an angle.
d/dx(arcsin x) = 1/√(1 − x²), |x| < 1
d/dx(arccos x) = −1/√(1 − x²), |x| < 1
d/dx(arctan x) = 1/(1 + x²), x real
Here arcsin means the inverse sine with output in [−π/2, π/2], arccos uses [0, π], and arctan uses (−π/2, π/2). All output angles and derivative formulas use radians.
The notation sin⁻¹x means arcsin x in this context, not 1/sin x. The reciprocal is cosec x and has a different derivative.
Principal branch and tangentExplore
arcsin x has domain [−1, 1] and range [−π/2, π/2].
x = 0; y = 0 radians; derivative = 1.
The derivative is finite at this interior input. Move towards an endpoint to compare slopes.
At an arcsin or arccos endpoint the function still has a value, but there is no finite derivative. The model shows a one-sided vertical tangent. Arctan has no finite domain endpoints.
02 / Derive the arcsine formula
The branch determines the sign of the square root.
Let y = arcsin x, so x = sin yWorked example
dx/dy = cos y
Differentiate the original sine function.
dy/dx = 1/cos y
Use the reciprocal-gradient rule at interior points.
cos y = √(1 − sin²y) = √(1 − x²)
cos y is positive when −π/2 < y < π/2.
dy/dx = 1/√(1 − x²), for −1 < x < 1
At x = ±1, cosine is zero and this finite derivative formula fails.
Watch arcsine’s tangent steepen towards an endpoint
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Derive the arccosine formula
The chosen branch makes sine positive.
Let y = arccos x, so x = cos yWorked example
dx/dy = −sin y
The original derivative has a minus sign.
sin y = √(1 − x²) for 0 < y < π
Sine is positive on the interior principal branch.
dy/dx = −1/√(1 − x²), |x| < 1
Arccosine decreases on its whole domain.
02 · A negative gradient
Find the derivative of arccos x at x = −1/2.
Hint
The denominator is the same positive square root as for x = 1/2.
Worked solution
The derivative is −2/√3. The function value is 2π/3 radians.
04 / Derive the arctangent formula
The denominator never becomes zero.
Let y = arctan x, so x = tan yWorked example
dx/dy = sec²y = 1 + tan²y
Use the Pythagorean identity.
dx/dy = 1 + x²
Substitute the original input back.
dy/dx = 1/(1 + x²), for every real x
Arctangent is increasing, with smaller gradients far from the origin.
03 · A scaled input
Differentiate arctan(5x).
Hint
Multiply by the inside derivative 5.
Worked solution
The derivative is 5/(1 + 25x²), valid for every real x.
05 / Include the inside derivative
Check the input range before applying the formula.
d/dx[arcsin g] = g′/√(1 − g²)
d/dx[arccos g] = −g′/√(1 − g²)
d/dx[arctan g] = g′/(1 + g²)
The first two formulas require |g| < 1 and a differentiable inside function. Endpoint cases need separate analysis.
Differentiate arcsin(x/3)Worked example
Original domain: −3 ≤ x ≤ 3
The arcsine input must lie between −1 and 1.
y′ = (1/3)/√(1 − x²/9)
Apply the chain rule for |x| < 3.
y′ = 1/√(9 − x²), |x| < 3
Use the positive constant 3 when simplifying the root.
04 · Shifted arccosine
Differentiate arccos(2x − 1). Give the original domain and derivative domain.
Hint
Solve −1 ≤ 2x − 1 ≤ 1.
Worked solution
The function domain is 0 ≤ x ≤ 1. Its derivative is −2/√[1 − (2x − 1)²], valid for 0 < x < 1. At either endpoint there is no finite derivative.
05 · A nonlinear input
Differentiate arccos(x²).
Hint
The inside derivative is 2x and the original input must satisfy x² ≤ 1.
Worked solution
The derivative is −2x/√(1 − x⁴), valid for −1 < x < 1. The function itself also exists at x = ±1.
06 / Differentiate further layers
A power of an inverse function is a composition.
Differentiate [arcsin x]²Worked example
u = arcsin x; y = u²
The last operation is squaring the angle.
y′ = 2u/√(1 − x²)
Use the chain rule.
y′ = 2arcsin x/√(1 − x²), |x| < 1
Keep the inverse function in the numerator.
06 · A root inside arctangent
Differentiate arctan(√x), and state where the formula is valid.
Hint
The original domain is x ≥ 0; the square root derivative requires x > 0.
Worked solution
The derivative is 1/[2√x(1 + x)] for x > 0. At zero the function has a value but its right-hand difference quotient is unbounded.
07 · Exponential inside an inverse
Differentiate arctan(eˣ).
Hint
The denominator becomes 1 + e²ˣ.
Worked solution
The derivative is eˣ/(1 + e²ˣ), for every real x.
07 / Use product and quotient rules around the inverse
Differentiate each part with the appropriate rule.
Differentiate x arctan xWorked example
u′ = 1; v′ = 1/(1 + x²)
Identify both changing factors.
y′ = arctan x + x/(1 + x²)
Add the two product contributions.
08 · A product with arcsine
Differentiate x²arcsin x.
Hint
The product has two terms.
Worked solution
The derivative is 2x arcsin x + x²/√(1 − x²), for |x| < 1.
09 · A quotient with a missing point
Differentiate (arctan x)/x on its original domain.
Hint
The numerator derivative is 1/(1 + x²).
Worked solution
The derivative is [x/(1 + x²) − arctan x]/x², for x ≠ 0. The original quotient is undefined at zero; a limiting value does not fill that point automatically.
08 / Do not cancel through a branch change
The square root of cos²x is |cos x|.
Differentiate arcsin(sin x) for 0 < x < πWorked example
For 0 < x < π/2: arcsin(sin x) = x
The angle is already in the principal arcsine range.
For π/2 < x < π: arcsin(sin x) = π − x
The inverse returns the reflected principal angle.
Derivatives: +1 before π/2, −1 after π/2
At π/2 the graph has a corner and is not differentiable.
Chain expression: cos x/√(1 − sin²x) = cos x/|cos x|
Replacing |cos x| by cos x would incorrectly give +1 on both sides.
10 · An inverse of a reciprocal input
Differentiate arctan(1/x). Is x = 0 in the original domain?
Hint
The inside derivative is −1/x².
Worked solution
The derivative simplifies to −1/(1 + x²), but only for x ≠ 0. The function jumps between limiting values −π/2 and +π/2 at zero, so the smooth-looking derivative formula does not extend the original function there.
11 · A branch test
Find the derivative of arcsin(sin x) at x = 3π/4.
Hint
This angle lies after the branch’s turning point π/2.
Worked solution
The derivative is −1. On this interval the function equals π − x, not x.
09 / Use exact angles for tangent questions
The derivative is a slope; the inverse value is an angle.
Tangent to y = arctan(2x) at x = 1/2Worked example
y = arctan 1 = π/4
The contact point is (1/2, π/4).
y′ = 2/(1 + 4x²); y′(1/2) = 1
Use the inside factor 2.
y − π/4 = x − 1/2
Keep the exact angle in the line equation.
12 · An arcsine normal
Find the normal to y = arcsin x at x = 1/2.
Hint
The point is (1/2, π/6) and the tangent slope is 2/√3.
Worked solution
The normal slope is −√3/2. Its equation is y − π/6 = −(√3/2)(x − 1/2).
13 · Stationary points
Find and classify all stationary points of y = arctan x − x/2.
Hint
Set 1/(1 + x²) − 1/2 = 0.
Worked solution
x = ±1. At x = −1, y = 1/2 − π/4, a minimum; at x = 1, y = π/4 − 1/2, a maximum. The derivative is negative for |x| > 1 and positive for |x| < 1.
10 / Extension: define inverse secant carefully
An explicit branch removes an ambiguity.
For this extension, define arcsec x = arccos(1/x). Its domain is x ≤ −1 or x ≥ 1, and its range is [0, π] excluding π/2. This is an inverse function, not the reciprocal of sec x.
Differentiate on |x| > 1Worked example
d/dx[arccos(1/x)] = 1/[x²√(1 − 1/x²)]
The negative arccos derivative and negative inner derivative cancel.
√(1 − 1/x²) = √(x² − 1)/|x|
The square root of x² is |x|, including on the negative branch.
d/dx(arcsec x) = 1/[|x|√(x² − 1)]
This derivative is positive on both open branches.
At x = ±1 the function exists but has no finite derivative. Always state a branch convention when using a less familiar inverse function.
14 · Negative branch
With this definition, find the derivative of arcsec x at x = −2.
Hint
The absolute value in the denominator is 2.
Worked solution
The derivative is 1/(2√3), which is positive. Omitting |x| would give the wrong sign.
15 · Composite inverse secant
Differentiate arcsec(2x) on the interior of its real domain.
Hint
Use the inside derivative 2 and the absolute value |2x|.
Worked solution
The derivative is 1/[|x|√(4x² − 1)] for |x| > 1/2. The function itself also exists at x = ±1/2.
11 / Extension: inverse cosecant and cotangent
Define the ranges before quoting formulas.
Define arccosec x = arcsin(1/x), with domain |x| ≥ 1 and range [−π/2, π/2] excluding zero. Then, for |x| > 1:
d/dx(arccosec x) = −1/[|x|√(x² − 1)]
Define arccot x = π/2 − arctan x, with domain all real x and range (0, π). It is continuous through x = 0 and has derivative:
d/dx(arccot x) = −1/(1 + x²)
Other branch conventions exist. These formulas and continuity statements refer to the definitions given here.
16 · Inverse cosecant
For the stated branch, find the derivative of arccosec x at x = −2.
Hint
The absolute value keeps the denominator positive.
Worked solution
The derivative is −1/(2√3). It is negative on both open branches.
17 · Inverse cotangent at zero
For arccot x = π/2 − arctan x, find the value and derivative at x = 0.
Hint
Use arctan 0 = 0.
Worked solution
The value is π/2 and the derivative is −1. This branch is defined and differentiable at zero.
12 / Check branch, domain and inside derivative
A valid value does not guarantee a finite derivative.
- Use the principal ranges and radians.
- For arcsin and arccos, distinguish the closed function domain from the open derivative domain.
- Multiply by the inside derivative for a composition.
- Keep absolute values when simplifying square roots of squares.
- Preserve holes and branch changes after simplification.
- For reciprocal inverse extensions, state the chosen branch explicitly.
Section 1 of 12 · Choose the principal inverse branch