Difficulty: Intermediate | Prerequisites: implicit differentiation, chain rule, basic trig identities, understanding of concavity.
This set of topics extends your differentiation toolkit beyond the standard rules. Local linearisation uses a tangent line to approximate a function's value near a known point, which is a technique that reappears throughout applied maths, physics and engineering. Inverse trigonometric derivatives give you the tools to differentiate expressions like arctan(2x) or arccos(x/2), which show up whenever angles are involved. The derivative-of-an-inverse-function formula rounds out the unit by connecting a function's derivative to its inverse's derivative, a relationship tested frequently in AP and university exams.
Linearisation replaces a curve with its tangent line near a known point to get quick approximations. Inverse trig derivatives follow memorisable formulas involving square roots and 1 + x² denominators. The derivative of an inverse function at a point is simply 1 over the derivative of the original function, evaluated at the corresponding point.
Local linearisation (linear approximation, tangent-line approximation)
The technique of using the tangent line at a convenient point a to estimate f(x) for x near a. Think of it as: the tangent line is a "good enough" stand-in for the curve when you are close to the point of tangency.
Overestimate vs underestimate
A linear approximation is an overestimate when the tangent line lies above the curve near the point, and an underestimate when it lies below. The sign of the second derivative tells you which: f''(a) > 0 (concave up) means the tangent is below the curve, so the approximation is an underestimate. f''(a) < 0 (concave down) means the tangent is above, so it is an overestimate.
Inverse trigonometric functions
The functions arcsin, arccos, arctan, arccot, arcsec, arccsc. Each "undoes" its corresponding trig function on a restricted domain. In simple terms, arctan(x) answers the question "what angle has tangent equal to x?"
Derivative of an inverse function
If f and g are inverses (so f(g(x)) = x), then g'(x) = 1 / f'(g(x)). In simple terms, the rate at which the inverse changes is the reciprocal of the rate at which the original function changes, evaluated at the right point.
The idea is simple: near a point x = a where you know f(a) and f'(a) exactly, the tangent line gives a good approximation of f(x).
The linearisation formula: L(x) = f(a) + f'(a)(x - a).
Example (Problem 6): approximate f(0.3) for f(x) = sqrt(9 + tan x), using x = 0 as the base point.
f(0) = sqrt(9 + 0) = 3.
f'(x) = sec²x / (2 sqrt(9 + tan x)). At x = 0: f'(0) = 1/6.
L(x) = 3 + (1/6)(x - 0) = (1/6)x + 3.
f(0.3) ≈ L(0.3) = (1/6)(0.3) + 3 = 3.05.
Once you have a linear approximation, exams often ask whether it is an overestimate or an underestimate. The answer comes from the second derivative at the base point.
If f''(a) > 0 (concave up), the curve bends upward away from the tangent line. The tangent sits below the curve, so the linearisation is an underestimate.
If f''(a) < 0 (concave down), the tangent sits above the curve, so the linearisation is an overestimate.
Example (Problem 7): g(2) = 1, g'(2) = 4, g''(2) = 3. Approximate g(1.9).
L(x) = 1 + 4(x - 2) = 4x - 7.
g(1.9) ≈ 4(1.9) - 7 = 0.6.
Since g''(2) = 3 > 0, the graph is concave up at x = 2, so the tangent line is below the curve and 0.6 is an underestimate.
These derivatives must be memorised. They appear frequently in both multiple-choice and free-response exam questions.
The key formulas (where u is a function of x and u' = du/dx):
d/dx [arcsin(u)] = u' / sqrt(1 - u²)
d/dx [arccos(u)] = -u' / sqrt(1 - u²)
d/dx [arctan(u)] = u' / (1 + u²)
d/dx [arccot(u)] = -u' / (1 + u²)
Notice the pattern: arcsin and arccos share the same denominator but differ in sign, and arctan and arccot share the same denominator but differ in sign.
Example (Problem 8): find d/dx [arccos(x/2)].
Here u = x/2, so u' = 1/2.
d/dx [arccos(x/2)] = -(1/2) / sqrt(1 - x²/4) = -1 / sqrt(4 - x²).
Answer: (B) -1/sqrt(4 - x²).
Example (Problem 9): find the tangent to f(x) = arctan(2x) at x = 0.
f(0) = arctan(0) = 0.
f'(x) = 2 / (1 + 4x²). At x = 0: f'(0) = 2.
Tangent: y = 2x. Answer: (C).
Example (Problem 10): given sin x = e^y, find dy/dx.
Differentiate implicitly: cos x = e^y · y'.
y' = cos x / e^y. Since e^y = sin x, this gives y' = cos x / sin x = cot x.
Answer: (D) dy/dx = cot x.
Example (Problem 12): slope of the tangent to f(x) = arccot(2x) at x = 1.
f'(x) = -2 / (1 + 4x²). At x = 1: f'(1) = -2/5.
Answer: (A) -2/5.
If f and g are inverse functions (f(g(x)) = x), the derivative of the inverse at a point is:
f'(x) = 1 / g'(f(x))
The key step is matching up the points correctly. If f(10) = 2, then g(2) = 10 on the original function's graph, and you evaluate g' at 2.
Example (Problem 13): g(x) = x³ + x, f(x) = g⁻¹(x), and f(10) = 2. Find f'(10).
Since f is the inverse of g, the pair (2, 10) on g corresponds to (10, 2) on f.
g'(x) = 3x² + 1. At x = 2: g'(2) = 13.
f'(10) = 1 / g'(2) = 1/13.
Answer: (B) 1/13.
Bonus from the source (Problem 11): d/dx [x⁵ - 5^x] = 5x⁴ - 5^x · ln 5. This uses the standard rules: the power rule for x⁵ and the exponential rule d/dx [a^x] = a^x · ln a. Answer: (D) 5x⁴ - (ln 5)5^x.
Linearisation: L(x) = f(a) + f'(a)(x - a)
Inverse trig derivatives (chain-rule form, u = u(x)):
d/dx [arcsin(u)] = u' / sqrt(1 - u²)
d/dx [arccos(u)] = -u' / sqrt(1 - u²)
d/dx [arctan(u)] = u' / (1 + u²)
d/dx [arccot(u)] = -u' / (1 + u²)
d/dx [arcsec(u)] = u' / (|u| sqrt(u² - 1))
d/dx [arccsc(u)] = -u' / (|u| sqrt(u² - 1))
Derivative of an inverse function: if f = g⁻¹, then f'(x) = 1 / g'(f(x))
Exponential derivative: d/dx [a^x] = a^x · ln a
Students often confuse the sign conventions for arcsin vs arccos derivatives. Arcsin has a positive numerator, arccos has a negative one. The denominators are the same.
When using the inverse function derivative formula, students frequently evaluate g' at the wrong point. If f(10) = 2, you need g'(2), not g'(10).
In linearisation problems, students sometimes pick a base point that is not easy to evaluate exactly. The whole method depends on choosing an a where f(a) and f'(a) can be computed without a calculator.
Students confuse overestimate and underestimate: concave up means the tangent is below the curve (underestimate), not above it.
⚠️ Inverse trig derivative formulas are tested as multiple-choice questions nearly every term. Know them cold.
⚠️ Linearisation questions almost always include a follow-up asking whether the approximation is an over- or underestimate. The second derivative test is the expected justification.
⚠️ The inverse function derivative formula (f'(x) = 1/g'(f(x))) is a favourite for its point-matching trap. Draw the two coordinate pairs on the two graphs if you need to keep them straight.
True or false: d/dx [arctan(x)] = 1 / (1 + x²). ___
Fill in the blank: if f''(a) > 0, the linear approximation at a is an ___.
True or false: d/dx [arccos(x)] = 1 / sqrt(1 - x²). ___
Fill in the blank: if f = g⁻¹ and f(b) = c, then f'(b) = 1 / g'(___). ___
True or false: d/dx [5^x] = 5^x · ln 5. ___
Answers: True; underestimate; False (it is negative: -1/sqrt(1 - x²)); c; True.
Q: Use linearisation to approximate sqrt(9 + tan(0.3)), using x = 0 as the base point.
A: f(0) = 3, f'(0) = 1/6. L(0.3) = 3 + (1/6)(0.3) = 3.05.
Q: g(2) = 1, g'(2) = 4, g''(2) = 3. Approximate g(1.9). Is this an over- or underestimate?
A: L(1.9) = 1 + 4(1.9 - 2) = 1 - 0.4 = 0.6. Since g''(2) > 0, the curve is concave up, so 0.6 is an underestimate.
Q: Find d/dx [arccos(x/2)].
A: -1/sqrt(4 - x²).
Q: Find the equation of the tangent line to f(x) = arctan(2x) at x = 0.
A: f(0) = 0, f'(0) = 2. Tangent: y = 2x.
Q: Given g(x) = x³ + x and f = g⁻¹ with f(10) = 2, find f'(10).
A: g'(x) = 3x² + 1. g'(2) = 13. f'(10) = 1/13.
Q: Differentiate x⁵ - 5^x.
A: 5x⁴ - 5^x · ln 5.
Linearisation is the first step toward Taylor and Maclaurin series (Calculus II), where you build higher-order polynomial approximations instead of just a straight line. Inverse trig derivatives reappear heavily in integration (Calculus II), because many integral forms resolve into arctan or arcsin expressions. The inverse function derivative formula connects to the broader idea of functional inverses and bijections, which matters in differential equations and real analysis.
linear approximation, local linearisation, tangent line approximation, overestimate, underestimate, concavity, second derivative test, inverse trig derivatives, arcsin derivative, arccos derivative, arctan derivative, arccot derivative, derivative of inverse function, inverse function theorem, exponential derivative, a^x derivative, ln a rule, Calculus I, MSU MTH 132, Unit 4