Differentiate inverse circular functions with exact domains

Start with principal values and radians

Use the principal branches: arcsin on [−1,1] with range [−π/2,π/2], arccos on [−1,1] with range [0,π], and arctan on R with range (−π/2,π/2). Apply the chain, product and quotient rules where needed. All angles here are in radians; an inverse function is not a reciprocal function

The inverse derivative is a reciprocal slope

At an interior point where a differentiable one-to-one function has nonzero derivative, its inverse derivative is the reciprocal slope at the corresponding point. For a direct justification, divide the inverse difference quotient Δy/Δx as1/(Δx/Δy). Continuity of the inverse makes Δy tend to0 with Δx. A nonzero limiting forward slope permits taking the reciprocal

Derive the arcsine rule

Write y=arcsin x, so x=sin y. On the interior principal interval, cos y>0. The reciprocal slope is1/cos y. Since cos²y=1−sin²y=1−x², the principal sign gives cos y=√(1−x²), hence dy/dx=1/√(1−x²) for−1<x<1. The inverse is defined at±1, but this finite derivative formula is not

Why is the square root positive in the arcsine derivation?

Derive the arccosine rule

Write y=arccos x, so x=cos y and dx/dy=−sin y. On0<y<π, sin y>0 and sin y=√(1−x²). Thus dy/dx=−1/√(1−x²) for−1<x<1. The negative sign agrees with the decreasing inverse-cosine graph

Derive the arctangent rule

Write y=arctan x, so x=tan y. The forward derivative is sec²y=1+tan²y=1+x², which is always positive. Thus dy/dx=1/(1+x²) for every real x. There is no square root and no finite endpoint exclusion

Give the derivative of arccos x at x=0.

Which derivative stays positive on all real inputs and approaches0 as |x| grows?

An affine input contributes its whole derivative

For f(x)=−2 arccos(3x−1)+4, the outer−2, inverse-cosine minus sign and inner derivative3 combine to give f′(x)=6/√[1−(3x−1)²]. The vertical shift differentiates to0. The original domain is[0,2/3] and the derivative domain is(0,2/3)

Enter the constant numerator in that derivative

Reverse inequalities when the inner slope is negative

For f(x)=3 arcsin(1−2x), require−1≤1−2x≤1, giving0≤x≤1. Its derivative is−6/√[1−(1−2x)²] on0<x<1. Sort the bounds after dividing by−2. The original endpoints remain defined even though the ordinary two-sided derivative is unavailable there

Give the upper endpoint of that original domain

For the nonconstant affine example, which statement is correct?

Evaluate an exact gradient before rounding

For f(x)=2 arcsin(3x−1), at x=1/2 the inverse input is1/2. Thus f′(1/2)=6/√(3/4)=12/√3=4√3. The exact answer can be entered as coefficient C=4 and square-free radicand r=3. For rational gradients use r=1.

Enter C in f′(1/2)=C√3 for that example

A quadratic input has a variable chain factor

For f(x)=−arccos[4(x−1)²], differentiate the inner input to8(x−1). The two minus signs cancel, giving f′(x)=8(x−1)/√[1−16(x−1)⁴]. The original domain is[1/2,3/2], and the derivative formula is valid on its interior. At x=1, the derivative is0 because its numerator is0 and its denominator is1.

When the numerator is written Mx+N, give N.

Construct a tangent or normal from the actual point

For f(x)=2 arccos(3x−1)+1, inverse input0 occurs at x₀=1/3. Since arccos0=π/2, y₀=π+1. The tangent gradient is−6 and the normal gradient is1/6. Their point forms are y=−6(x−1/3)+π+1 and y=(x−1/3)/6+π+1. Keeping theπ coefficient exact avoids rounding the intercept

Give the normal gradient in that example

Combine inverse rules with their separate denominators

For f(x)=3 arcsin(2x)+arccos(2x)+2 arctan(2x), combine the first two derivatives to obtain4/√(1−4x²), then add4/(1+4x²). The original domain is[−1/2,1/2], and the ordinary derivative domain is its interior. The arctangent term does not remove the other terms’ input restriction

Give the numerator over√(1−4x²) in that derivative

Check stationary candidates before accepting squared equations

For f(x)=20 arcsin x−34 arctan x, stationary points require10/√(1−x²)=17/(1+x²), with−1<x<1. Both sides are positive, so squaring is reversible. With y=x², the equation becomes100y²+489y−189=0, or(25y−9)(4y+21)=0. Reject y=−21/4. The remaining y=9/25 gives x=±3/5; both satisfy the original unsquared equation

Give the positive stationary x-value in that example

Derive all three inverse rules, ungraded

Use reciprocal difference quotients to justify the inverse derivative rules. State the relevant principal intervals, explain each square-root sign and identify the open derivative domains. Distinguish this argument from merely rearranging a chain-rule equation after assuming a derivative exists

Model: justify the limiting reciprocal

For each continuous strictly monotone principal branch, its inverse is continuous. At an interior point, Δy/Δx=1/(Δx/Δy), and Δy tends to0 as Δx does. The forward quotient tends to cos y₀,−sin y₀ or sec²y₀ respectively. Each is nonzero on its interior branch, so the reciprocal limit exists. Their signs and identities give1/√(1−x₀²),−1/√(1−x₀²) and1/(1+x₀²). The first two arguments require−1<x₀<1; arctan works for every real x₀

Inspect an endpoint composition separately, ungraded

Explain why the standard arcsin chain formula requires |u(x)|<1, but its failure to apply does not prove that every composition with u(x)=1 is nondifferentiable. Give a counterexample. Also explain why cancelling equal arcsin/arccos contributions does not enlarge the original domain

Model: formula applicability is not a universal nonexistence test

If u(x)=1 for every real x, then arcsin(u(x))=π/2 is constant and has derivative0 everywhere. The formal quotient u′/√(1−u²) is0/0, so that formula cannot establish the result; the constant-function definition does. For arcsin x+arccos x, the original domain is still[−1,1], even though its values equal the constantπ/2. Extending that constant to R defines a different function. The present affine and quadratic practice families avoid such ambiguous endpoint analyses by stating their exact original domains

Use product and quotient rules, ungraded

Differentiate x arctan x and(arcsin x)/x on their original real domains. Show where both the inverse and the ordinary product/quotient rules enter. State the derivative domains rather than silently filling the hole at x=0.

Model: retain every original restriction

The product derivative is arctan x+x/(1+x²) on R. The quotient derivative is[x/√(1−x²)−arcsin x]/x² on(−1,0)∪(0,1). The original quotient has domain[−1,0)∪(0,1]; it is undefined at0. A continuous extension or a limiting value at0 is a separate construction, not an input restored by differentiation

Check numerically with technology, ungraded

In radian mode, compare the central difference for f(x)=−2 arccos(3x−1)+4 at x=1/3 with the exact derivative. Use several decreasing positive increments that keep both evaluation points inside(0,2/3). Then explain why a central difference at an original domain endpoint is not permitted. Record your own calculations or code; supplied reference values do not establish that you executed the check

Model: finite differences support the exact result

At x=1/3, the derivative is6. With increment0.01, the central difference is approximately6.0009, approaching6 as the increment shrinks. Both inputs must lie in the original domain. At x=0 or2/3, one side of a symmetric increment lies outside it. Floating-point differences near a singular boundary may be unreliable; the exact domain and symbolic derivation determine the result

Exact values and written explanations

Find exact derivative coefficients, gradients, domain bounds, lines and stationary x-values in the stated forms. General symbolic expressions, independent derivations, complete written proofs and technology execution remain ungraded