Inverse functions and the inverse trig derivatives · Calculus I: Limits & Derivatives

Last lesson differentiated \ln x by writing e^{\ln x} = x and differentiating both sides. That wasn't a trick specific to logarithms — it's the general method for inverses, and it produces the entire inverse-trig table.

The rule

If f is differentiable and one-to-one, and g = f^{-1}, then

$g'(x) = \frac{1}{f'(g(x))}$

Derivation. By definition f(g(x)) = x. Differentiate both sides, chain rule on the left:

$f'(g(x))\cdot g'(x) = 1 \implies g'(x) = \frac{1}{f'(g(x))}$

The only requirement is f'(g(x)) \neq 0 — and where it is zero, f has a horizontal tangent, so its inverse has a vertical one and isn't differentiable there. The formula reports that honestly by dividing by zero.

Why the reciprocal

The graph of f^{-1} is the graph of f reflected across y = x (§0.2). Reflection swaps the horizontal and vertical axes, so it swaps rise and run — and a slope of \frac{\text{rise}}{\text{run}} becomes \frac{\text{run}}{\text{rise}}.

That's the whole content: reflecting a graph inverts its slopes.

The point-form is worth stating separately because it's what you actually compute with:

$(f^{-1})'(b) = \frac{1}{f'(a)} \quad\text{where } f(a) = b$

Find the matching input first, evaluate f' there, reciprocate. You never need a formula for f^{-1} — which is the point, because usually there isn't one.

Example without a formula. Let f(x) = x^5 + 2x + 1. It's strictly increasing (its derivative 5x^4+2 is always positive), so an inverse exists, but solving y = x^5+2x+1 for x is hopeless.

Still: f(1) = 4, so f^{-1}(4) = 1, and

$(f^{-1})'(4) = \frac{1}{f'(1)} = \frac{1}{5+2} = \frac17$

No formula for the inverse required.

The inverse trig functions

Sine isn't one-to-one, so \arcsin is defined by restricting the domain to \left[-\frac\pi2, \frac\pi2\right], where sine increases from -1 to 1. Similar restrictions define the others. Those choices are conventions, but universal ones.

Deriving \frac{d}{dx}\arcsin x. Let y = \arcsin x, so \sin y = x with y \in \left[-\frac\pi2,\frac\pi2\right]. Differentiate both sides:

$\cos y\cdot\frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos y}$

Now convert \cos y into something in terms of x. From \sin^2 y + \cos^2 y = 1:

$\cos y = \pm\sqrt{1 - \sin^2 y} = \pm\sqrt{1-x^2}$

The sign is decided by the domain restriction. On \left[-\frac\pi2,\frac\pi2\right] cosine is non-negative, so take the positive root:

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$

That last step is where the restriction earns its keep. Without it the derivative would be ambiguous.

\arctan, the same way. \tan y = x gives \sec^2y\cdot y' = 1, and \sec^2 y = 1 + \tan^2 y = 1+x^2, so

$\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$

No sign ambiguity at all here, since \sec^2 is never negative. This is the tidiest of the six.

The table

f(x)	f'(x)	Domain
\arcsin x	\dfrac{1}{\sqrt{1-x^2}}	(-1,1)
\arccos x	-\dfrac{1}{\sqrt{1-x^2}}	(-1,1)
\arctan x	\dfrac{1}{1+x^2}	all \mathbb R
\operatorname{arccot} x	-\dfrac{1}{1+x^2}	all \mathbb R
\operatorname{arcsec} x	\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}}	\lvert x\rvert>1
\operatorname{arccsc} x	-\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}}	\lvert x\rvert>1

Two observations that make this memorable:

The "co-" versions are just negatives. That's not a coincidence: \arcsin x + \arccos x = \frac\pi2 identically, so differentiating gives \arcsin' + \arccos' = 0. Same for the other two pairs. Three formulas, not six.

Algebraic derivatives from transcendental functions. \arctan is deeply non-algebraic, and its derivative is \frac{1}{1+x^2}. Read backwards that says

$\int\frac{dx}{1+x^2} = \arctan x, \qquad \int\frac{dx}{\sqrt{1-x^2}} = \arcsin x$

which is why inverse trig functions appear all over §4 as answers to integrals that contain no trigonometry whatsoever. It's also the source of the classic series \frac\pi4 = 1 - \frac13 + \frac15 - \cdots in §8.3.

With the chain rule

$\frac{d}{dx}\arctan(u) = \frac{u'}{1+u^2}, \qquad \frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1-u^2}}$

$\frac{d}{dx}\arctan(3x) = \frac{3}{1+9x^2}, \qquad \frac{d}{dx}\arcsin(x^2) = \frac{2x}{\sqrt{1-x^4}}$

Doing it in Python

The reciprocal rule, without ever forming the inverse:

from math import sqrt

def numerical(f, x, h=1e-6):
    return (f(x + h) - f(x - h)) / (2 * h)

f = lambda t: t**5 + 2*t + 1
fp = lambda t: 5*t**4 + 2

# f(1) = 4, so f_inverse(4) = 1
def f_inverse(y, lo=-5.0, hi=5.0):
    for _ in range(200):
        mid = (lo + hi) / 2
        lo, hi = (mid, hi) if f(mid) < y else (lo, mid)
    return (lo + hi) / 2

print(f"f(1) = {f(1)}, so f_inverse(4) = {f_inverse(4):.10f}")
print(f"rule    (f_inv)'(4) = 1/f'(1) = {1/fp(1):.10f}")
print(f"numeric (f_inv)'(4) = {numerical(f_inverse, 4.0, h=1e-5):.10f}")

The whole inverse-trig table:

from math import asin, acos, atan, sqrt

def numerical(f, x, h=1e-6):
    return (f(x + h) - f(x - h)) / (2 * h)

table = [
    ("arcsin", asin, lambda x: 1 / sqrt(1 - x*x)),
    ("arccos", acos, lambda x: -1 / sqrt(1 - x*x)),
    ("arctan", atan, lambda x: 1 / (1 + x*x)),
]

x = 0.4
print(f"{'f':>8} {'rule':>14} {'numerical':>14}")
for name, f, df in table:
    print(f"{name:>8} {df(x):>14.8f} {numerical(f, x):>14.8f}")

print(f"\narcsin(x) + arccos(x) = {asin(x) + acos(x):.10f} = pi/2, always")
print("so their derivatives must sum to zero -- which is why one is the other's negative")

The reflection picture, in numbers:

from math import exp, log

print(f"{'x':>6} {'e^x':>10} {'slope of e^x':>14} "
      f"{'point on ln':>14} {'slope of ln':>14}")
for x in (-1.0, 0.0, 1.0, 2.0):
    y = exp(x)
    print(f"{x:>6} {y:>10.5f} {y:>14.5f} {'(' + format(y, '.3f') + ', ' + str(x) + ')':>14} "
          f"{1/y:>14.5f}")

print("\nreflecting swaps the axes, so rise/run becomes run/rise: the slopes reciprocate")

Where the rule breaks:

from math import sqrt

# f(x) = x^3 has f'(0) = 0, so its inverse has a vertical tangent at 0
def cbrt(y):
    return y ** (1/3) if y >= 0 else -((-y) ** (1/3))

print(f"{'y':>12} {'slope of cbrt':>16} {'1/(3 x^2) at x=cbrt(y)':>26}")
for y in (1.0, 1e-2, 1e-4, 1e-6):
    x = cbrt(y)
    print(f"{y:>12.0e} {(cbrt(y + 1e-9) - cbrt(y - 1e-9)) / 2e-9:>16.2f} "
          f"{1 / (3 * x**2):>26.2f}")

print("\nf'(0) = 0 means the inverse's slope blows up -- a vertical tangent (2.1)")

Worked example

Differentiate y = \arctan\!\left(\sqrt{x}\right), and find the slope at x = 1.

Chain rule with u = \sqrt x:

$\frac{dy}{dx} = \frac{1}{1+u^2}\cdot u' = \frac{1}{1+(\sqrt x)^2}\cdot\frac{1}{2\sqrt x} = \frac{1}{2\sqrt x\,(1+x)}$

At x = 1: \frac{1}{2\cdot1\cdot2} = \frac14.

Sanity checks. As x \to 0^+ the slope blows up — and it should, since \sqrt x has a vertical tangent at 0 and composing preserves that. As x \to \infty the slope \to 0, matching \arctan flattening toward \frac\pi2.

A second one, showing where the domain bites. Differentiate y = \arcsin(2x):

$\frac{dy}{dx} = \frac{2}{\sqrt{1-4x^2}}$

The original function needs |2x| \le 1, i.e. |x| \le \frac12; the derivative needs the strict version |x| < \frac12. At exactly x = \pm\frac12 the function is defined but has a vertical tangent. Inverse functions inherit their domain restrictions, and the derivative's domain is usually one notch smaller than the function's.

Your turn

1. f(x) = x^3 + x + 1. Find (f^{-1})'(3).

2. \dfrac{d}{dx}\arctan(x^2)

3. \dfrac{d}{dx}\left[x\arcsin x + \sqrt{1-x^2}\right]

4. Why is \frac{d}{dx}\operatorname{arcsec} x = \frac{1}{|x|\sqrt{x^2-1}} rather than without the absolute value?

Solutions

1. Find the input mapping to 3: try x=1, giving 1+1+1 = 3 ✓. So f^{-1}(3) = 1.

$f'(x) = 3x^2+1 \implies f'(1) = 4$

$(f^{-1})'(3) = \frac{1}{f'(1)} = \boxed{\frac14}$

2. Chain rule with u = x^2:

$\frac{2x}{1+(x^2)^2} = \boxed{\frac{2x}{1+x^4}}$

3. Product rule on the first term, chain rule on the second:

$\frac{d}{dx}\left[x\arcsin x\right] = \arcsin x + \frac{x}{\sqrt{1-x^2}}$

$\frac{d}{dx}\sqrt{1-x^2} = \frac{-2x}{2\sqrt{1-x^2}} = \frac{-x}{\sqrt{1-x^2}}$

Adding, the two fractional terms cancel exactly:

$\boxed{\arcsin x}$

That cancellation is not luck — it means x\arcsin x + \sqrt{1-x^2} is the antiderivative of \arcsin x, which is how §4.6 computes \int\arcsin x\,dx by parts.

4. Because \operatorname{arcsec} is increasing on both branches of its domain, so its derivative must be positive for x < -1 as well as for x > 1.

Deriving it: \sec y = x gives \sec y\tan y\cdot y' = 1, so y' = \frac{1}{\sec y\tan y} = \frac{1}{x\tan y}. Then \tan y = \pm\sqrt{\sec^2y - 1} = \pm\sqrt{x^2-1}, and the sign depends on which branch you're on.

Working the cases through, both give a positive result, and \frac{1}{|x|\sqrt{x^2-1}} is the single expression covering both. Drop the absolute value and you get a negative derivative for x<-1, contradicting the graph.

This is exactly the \sqrt{x^2} = |x| care from §1.6, reappearing.

Check yourself in code

Verify the inverse-function rule for four functions.

For \arcsin at x=0.6, \arctan at x=0.6, \ln at x=2, and \sqrt[3]{x} at x=8: print the central difference (h=10^{-6}) and the formula's value, each to 8 decimals.

Print exactly this:

arcsin   x=0.6   numeric 1.25000000  rule 1.25000000
arctan   x=0.6   numeric 0.73529412  rule 0.73529412
ln       x=2.0   numeric 0.50000000  rule 0.50000000
cbrt     x=8.0   numeric 0.08333333  rule 0.08333333

from math import asin, atan, log, sqrt

def numerical(f, x, h=1e-6):
    return (f(x + h) - f(x - h)) / (2 * h)

cases = [
    ("arcsin", asin, lambda x: 1 / sqrt(1 - x*x), 0.6),
    ("arctan", atan, lambda x: 1 / (1 + x*x), 0.6),
    ("ln", log, lambda x: 1 / x, 2.0),
    ("cbrt", lambda t: t ** (1/3), lambda x: 1 / (3 * x**(2/3)), 8.0),
]

for name, f, df, x in cases:
    # print the numerical derivative and the rule's value
    print(f"{name:<8} x={x:<5} ...")

from math import asin, atan, log, sqrt

def numerical(f, x, h=1e-6):
    return (f(x + h) - f(x - h)) / (2 * h)

cases = [
    ("arcsin", asin, lambda x: 1 / sqrt(1 - x*x), 0.6),
    ("arctan", atan, lambda x: 1 / (1 + x*x), 0.6),
    ("ln", log, lambda x: 1 / x, 2.0),
    ("cbrt", lambda t: t ** (1/3), lambda x: 1 / (3 * x**(2/3)), 8.0),
]

for name, f, df, x in cases:
    print(f"{name:<8} x={x:<5} numeric {numerical(f, x):.8f}  rule {df(x):.8f}")

(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}, because reflecting a graph across y = x turns rise-over-run into run-over-rise. You never need a formula for the inverse — just the matching input. Applied to the restricted trig functions it gives \arcsin' = \frac{1}{\sqrt{1-x^2}} and \arctan' = \frac{1}{1+x^2}, with the "co-" versions differing only by a sign. Those algebraic derivatives are why inverse trig functions turn up as answers throughout §4.

Next: the hyperbolic functions, which mirror all of this with the signs changed — and one of which you already know as an activation function.

23. Inverse functions and the inverse trig derivatives