Difficulty: Intermediate | Prerequisites: Basic differentiation (power rule, chain rule, product rule)
This material covers how to use derivatives to locate and classify the turning points of a function, and how to handle equations where y is not isolated. These techniques feed directly into optimisation and related-rates problems later in the course. If you are comfortable taking first derivatives of polynomials and exponentials, you have the foundation.
Critical points occur where the first derivative is zero or undefined. The second derivative test tells you whether a critical point is a local maximum (concave down) or minimum (concave up). Implicit differentiation lets you find dy/dx when y is tangled with x in an equation, and marginal cost is simply the derivative of the cost function.
Critical point
A value x = c where f(c) is defined and either f'(c) = 0 or f'(c) does not exist. In simple terms, it is any point where the slope of the curve is flat or has a sharp corner.
Second Derivative Test
A method for classifying critical points. If f'(c) = 0 and f''(c) > 0, the point is a relative minimum. If f'(c) = 0 and f''(c) < 0, the point is a relative maximum. If f''(c) = 0, the test is inconclusive. Think of it as checking whether the curve bends upward or downward at the flat spot.
Implicit differentiation
A technique for finding dy/dx when y is not written as an explicit function of x. You differentiate both sides of the equation with respect to x, apply the chain rule to any y terms (tacking on dy/dx), and then solve for dy/dx.
Marginal cost
The derivative of the total cost function, C'(x). It represents the approximate cost of producing one additional unit. In simple terms, it tells you how much your costs go up when you make one more item.
Concavity
The direction a curve bends. Concave up means the curve opens upward (like a cup), and the second derivative is positive. Concave down means it opens downward (like a hill), and the second derivative is negative.
Product rule
The differentiation rule for products of two functions: if f(x) = u(x) · v(x), then f'(x) = u'v + uv'. Think of it as "derivative of the first times the second, plus the first times the derivative of the second."
Take the first derivative f'(x).
Set f'(x) = 0 and solve for x. Also note any x-values where f'(x) is undefined but f(x) is defined.
Those x-values are the critical points.
Example: f(x) = x^3 - 6x^2 + 1. f'(x) = 3x^2 - 12x = 3x(x - 4). Setting f'(x) = 0 gives x = 0 and x = 4.
Once you have a critical point where f'(c) = 0, compute f''(c).
f''(c) > 0: the curve is concave up at c, so c is a relative minimum.
f''(c) < 0: the curve is concave down at c, so c is a relative maximum.
f''(c) = 0: the test tells you nothing. Use the first derivative test or another method.
Example: f(x) = x^3 - 6x^2 + 9x + 1. f'(x) = 3x^2 - 12x + 9. Critical points at x = 1 and x = 3. f''(x) = 6x - 12. f''(1) = -6 < 0 (relative max). f''(3) = 6 > 0 (relative min).
The second derivative f''(x) is the derivative of f'(x). Apply the same differentiation rules a second time.
Example: g(t) = e^(3t). g'(t) = 3e^(3t). g''(t) = 9e^(3t).
Example: f(u) = u + u^2 + 5. f'(u) = 1 + 2u. f''(u) = 2.
If f(x) = u(x) · v(x), then f'(x) = u'(x) · v(x) + u(x) · v'(x).
Example: f(x) = x · ln(x). Let u = x, v = ln(x). u' = 1, v' = 1/x. f'(x) = (1)(ln x) + (x)(1/x) = ln(x) + 1.
Differentiate every term with respect to x. Whenever you differentiate a y-term, multiply by dy/dx (chain rule).
Collect all dy/dx terms on one side and solve.
Example: y^2 + 3xy = 2. Differentiating: 2y(dy/dx) + 3y + 3x(dy/dx) = 0. Factor: (2y + 3x)(dy/dx) = -3y. So dy/dx = -3y / (2y + 3x).
Example: y^2 = x^3 + 1. Differentiating: 2y(dy/dx) = 3x^2. So dy/dx = 3x^2 / (2y). At the point (2, 3): dy/dx = 12/6 = 2. Tangent line: y - 3 = 2(x - 2).
Marginal cost is C'(x), the first derivative of the cost function.
Example: C(x) = 0.5x^2 + 10x + 50. C'(x) = x + 10.
f'(c) = 0 \text{ or undefined} \Rightarrow x = c \text{ is a critical point (provided } f(c) \text{ is defined)}\text{Product Rule: } (uv)' = u'v + uv'\text{Implicit: } \frac{d}{dx}[y^2] = 2y\frac{dy}{dx}\text{Marginal cost} = C'(x)Students often think a critical point must be a maximum or minimum. A critical point is simply where f' = 0 or is undefined; it might be neither a max nor a min (for example, an inflection point with a horizontal tangent).
Students forget that a critical point also requires f(c) to be defined. If the function itself is undefined at c, that value is not a critical point.
In implicit differentiation, students differentiate y-terms as if y were a constant, forgetting the dy/dx factor from the chain rule.
Students sometimes confuse marginal cost with average cost. Marginal cost is C'(x), the derivative; average cost is C(x)/x.
⚠️ The definition of a critical point (f' = 0 OR f' undefined, with f defined) is tested directly in multiple choice. Know the precise wording.
⚠️ The second derivative test is a frequent short-answer question. Be ready to find critical points, then classify them.
⚠️ Implicit differentiation appears in both stand-alone questions and as a step inside related-rates and tangent-line problems. Practise the product rule on mixed xy terms.
⚠️ Marginal cost is a one-step question: differentiate the cost function. It is almost always on the exam.
True or False: A critical point of f(x) occurs only where f'(x) = 0. (False. It also occurs where f'(x) is undefined, provided f(x) is defined.)
Fill in the blank: If f'(c) = 0 and f''(c) > 0, then x = c is a relative ______. (minimum)
True or False: In implicit differentiation, the derivative of y^2 with respect to x is 2y. (False. It is 2y · dy/dx.)
Fill in the blank: The marginal cost of C(x) = 0.5x^2 + 10x + 50 is ______. (x + 10)
True or False: The second derivative of e^(3t) is 6e^(3t). (False. It is 9e^(3t).)
Q: What condition is required for x = c to be a critical point of g(x)?
A: g(c) must be defined, and either g'(c) = 0 or g'(c) is undefined.
Q: Given f(x), if f'(c) = 0 and f''(c) > 0, what can be concluded about x = c?
A: It is a relative minimum (second derivative test, concave up).
Q: If f(x) = x · ln(x), what is f'(x)?
A: ln(x) + 1, by the product rule.
Q: Use implicit differentiation to find dy/dx for y^2 + 3xy = 2.
A: dy/dx = -3y / (2y + 3x).
Q: Find the marginal cost if C(x) = 0.5x^2 + 10x + 50.
A: C'(x) = x + 10.
Q: Find the critical points of f(x) = x^3 - 6x^2 + 1.
A: f'(x) = 3x^2 - 12x = 3x(x - 4) = 0 gives x = 0 and x = 4.
Q: Calculate the second derivative of g(t) = e^(3t).
A: g'(t) = 3e^(3t), g''(t) = 9e^(3t).
Critical points and the second derivative test are the foundation of optimisation problems (maximising area, revenue, or profit under constraints). Implicit differentiation is essential for related-rates problems, where you differentiate a geometric or physical relationship with respect to time. Marginal cost connects to marginal revenue and elasticity of demand in the economics-application questions.
critical point, critical number, first derivative test, second derivative test, concavity, concave up, concave down, inflection point, product rule, implicit differentiation, dy/dx, marginal cost, marginal revenue, chain rule, MAT 210, brief calculus, ASU, relative max, relative min, local extrema