01 · Two differentiations
Find y′ and y″ for y = 3x⁴ − 2x³ + 5x − 7.
Hint
Differentiate each polynomial term twice.
Worked solution
y′ = 12x³ − 6x² + 5; y″ = 36x² − 12x.
Understand · explore · practise
Second derivatives, concavity and convexity
Calculate second derivatives and distinguish increasing or decreasing functions from concave or convex curves. Use sign intervals, repeated chain and product rules, and implicit or parametric second derivatives.
Before you startFirst derivatives; chain, product and quotient rules; implicit and parametric differentiation
01 / Differentiate the gradient
The second derivative measures how the first derivative changes.
f″(x) = d/dx[f′(x)]
d²y/dx² = d/dx(dy/dx)
The second derivative is a second differentiation, not the square of the first derivative. If f″ is positive on an interval, the tangent gradients increase as x increases. The function itself could still be decreasing.
Slope and change of slopeExplore
f′ = 2x; f″ = 2.
x = −1; f = 1; f′ = −2; f″ = 2.
The function is decreasing at this point.
The tangent slope is increasing here: the curve bends upwards.
A sign for f′ describes rising or falling. A sign for f″ describes whether the slope is increasing or decreasing. A zero second derivative at one point needs a check on either side.
02 / Differentiate twice carefully
Write the first derivative before finding the second.
y = x⁴ − 4x³ + 2xWorked example
y′ = 4x³ − 12x² + 2
Differentiate the original function.
y″ = 12x² − 24x = 12x(x − 2)
Differentiate the first derivative.
At x = 1: y′ = −6 and y″ = −12
The function is decreasing and its gradient is becoming more negative.
Watch a negative slope become less negative
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Second derivative is not a square
For y = x³, compare y″ and (y′)² at x = 1.
Hint
y′ = 3x².
Worked solution
y″ = 6x gives 6, while (y′)² = 9x⁴ gives 9. These are different operations.
03 / Connect the signs to curve shape
State the direction of bending as well as the terminology.
f″ > 0: slope increasing; concave up / convex
f″ < 0: slope decreasing; concave down
Some courses use “concave” to mean concave down and “convex” to mean concave up. Writing the sign and direction of bending avoids ambiguity.
For a twice-differentiable function on an interval, f″ ≥ 0 throughout is sufficient for convexity; f″ ≤ 0 throughout is sufficient for concavity downwards. An isolated zero does not automatically split a convex interval.
03 · Read a sign statement
On an interval, f′ < 0 and f″ > 0. Describe the graph.
Hint
The two derivative signs answer different questions.
Worked solution
The function is decreasing, but its gradients are increasing: negative slopes become less negative. The curve is convex, or concave up.
04 / Separate the four sign combinations
Rising and bending upwards are not the same condition.
Simple examples at x = 1Worked example
f = x²: f′ = 2, f″ = 2
Increasing and bending upwards.
f = −x²: f′ = −2, f″ = −2
Decreasing and bending downwards.
f = e⁻ˣ: f′ = −e⁻¹, f″ = e⁻¹
Decreasing and bending upwards.
f = ln x: f′ = 1, f″ = −1
Increasing and bending downwards.
04 · An increasing concave curve
For f(x) = ln x, state the signs of f′ and f″ over its real domain.
Hint
The domain is x > 0.
Worked solution
f′ = 1/x > 0 and f″ = −1/x² < 0. The curve is increasing and concave down everywhere on (0, ∞).
05 · A decreasing convex curve
Find the first two derivatives of f(x) = e^(−3x) and describe its shape.
Hint
Each differentiation supplies a factor −3.
Worked solution
f′ = −3e^(−3x) < 0 and f″ = 9e^(−3x) > 0. It decreases and bends upwards for all real x.
05 / Find intervals from a factored second derivative
Use zeros and exclusions to divide the number line.
For y = x⁴ − 4x³ + 2x, y″ = 12x(x − 2)Worked example
Potential sign boundaries: x = 0 and x = 2
There are no domain exclusions.
For x < 0: both factors are negative, so y″ > 0
The curve is convex on this interval.
For 0 < x < 2: the factors have opposite signs, so y″ < 0
The curve is concave down here.
For x > 2: both factors are positive, so y″ > 0
The curve is convex again.
Using the non-strict convexity criterion, the closed boundary points may be included in the adjacent intervals: y″ ≥ 0 on (−∞, 0] and [2, ∞), and y″ ≤ 0 on [0, 2]. The open intervals describe where each sign is strict.
06 · A cubic’s shape
Find where y = x³ − 3x² + 4 is concave up and concave down.
Hint
y″ = 6x − 6.
Worked solution
It is concave down for x < 1 and concave up for x > 1. The second derivative is zero at x = 1; its sign changes there.
06 / A zero second derivative needs context
Compare x³ with x⁴ at the origin.
Two zero second derivativesWorked example
For x³: f″ = 6x
The sign changes from negative to positive at zero.
For x⁴: f″ = 12x²
It is positive on both sides and zero only at the origin.
x⁴ is convex across zero
The isolated zero does not create a concavity change.
07 · Convexity with an isolated zero
Is y = x⁶ convex on the whole real line? Does y″ change sign at zero?
Hint
y″ = 30x⁴.
Worked solution
Yes, it is convex because y″ ≥ 0 everywhere. The second derivative is positive away from zero and does not change sign there.
08 · A straight line
For y = 5x − 2, find y″ and interpret the shape.
Hint
The first derivative is the constant 5.
Worked solution
y″ = 0 everywhere: the slope does not change. Under the non-strict definitions a straight line is both convex and concave. It has no change of concavity.
07 / Repeat the chain and product rules when needed
A correct first derivative may still have several changing factors.
y = e^(x²)Worked example
y′ = 2x e^(x²)
Apply the chain rule.
y″ = 2e^(x²) + 4x²e^(x²)
Use the product rule and chain rule again.
y″ = (2 + 4x²)e^(x²) > 0
The curve is convex for all real x.
09 · A product twice
Find y″ if y = x eˣ.
Hint
First y′ = eˣ(x + 1).
Worked solution
y″ = eˣ(x + 2). Its sign is negative for x < −2 and positive for x > −2.
10 · A reciprocal twice
Find the first two derivatives of y = 1/(x + 1), with x ≠ −1.
Hint
Write the function as (x + 1)⁻¹.
Worked solution
y′ = −1/(x + 1)² and y″ = 2/(x + 1)³. The curve is concave down for x < −1 and convex for x > −1. There is no curve point at x = −1.
08 / Keep the inner-angle factors at every differentiation
All angles in these formulas use radians.
y = sin(3x)Worked example
y′ = 3 cos(3x)
The first chain-rule factor is 3.
y″ = −9 sin(3x)
The second differentiation supplies another factor 3 and a minus sign.
11 · A cosine twice
Find y″ for y = 2 cos(4x − 1).
Hint
First y′ = −8 sin(4x − 1).
Worked solution
y″ = −32 cos(4x − 1).
09 / Differentiate the implicit relation again
Terms containing y′ also change with x.
x² + y² = 25, with y ≠ 0Worked example
2x + 2yy′ = 0
First implicit differentiation.
2 + 2(y′)² + 2yy″ = 0
Use the product rule on 2yy′.
y″ = −[1 + (y′)²]/y
Collect the second derivative.
y″ = −(x² + y²)/y³ = −25/y³
Substitute y′ = −x/y and use the original circle equation.
12 · Circle branch curvature
Find y″ at (3, 4) and (3, −4) on x² + y² = 25.
Hint
Use y″ = −25/y³.
Worked solution
At (3, 4), y″ = −25/64, so the upper branch bends downwards. At (3, −4), y″ = 25/64, so the lower branch bends upwards.
10 / A parametric second derivative needs another division
Differentiate with respect to x, not just the parameter.
d²y/dx² = [d/dt(dy/dx)]/(dx/dt), when dx/dt ≠ 0
x = t² + 1, y = t³, with t ≠ 0Worked example
dy/dx = (3t²)/(2t) = 3t/2
Find the first derivative.
d/dt(dy/dx) = 3/2
This is a derivative with respect to t.
d²y/dx² = (3/2)/(2t) = 3/(4t)
Divide by dx/dt to differentiate with respect to x.
13 · Check a parametric second derivative
For x = eᵗ, y = t², find d²y/dx².
Hint
First dy/dx = 2t e⁻ᵗ.
Worked solution
d/dt(dy/dx) = 2(1 − t)e⁻ᵗ. Divide by eᵗ to obtain d²y/dx² = 2(1 − t)e^(−2t), for every real t.
11 / Use the second derivative at a stationary point
The test first requires y′ = 0.
At a stationary point:
y″ > 0 → local minimum
y″ < 0 → local maximum
y″ = 0 → this test is inconclusive
y = x³ − 3xWorked example
y′ = 3x² − 3 = 0 gives x = ±1
Find all stationary inputs.
y″ = 6x
Evaluate it at those inputs.
At (−1, 2), y″ = −6: local maximum
The gradients decrease through zero.
At (1, −2), y″ = 6: local minimum
The gradients increase through zero.
14 · An inconclusive test
Classify the stationary point of y = x⁴ at zero even though y″(0) = 0.
Hint
Inspect y′ = 4x³ on either side.
Worked solution
The derivative changes from negative to positive, so (0, 0) is a minimum. A zero second derivative did not rule out a minimum.
15 · Do not skip the stationary condition
At x = 1, y = eˣ has y″ > 0. Is this a local minimum?
Hint
Check y′ as well.
Worked solution
No. y′ = eˣ is positive everywhere, so the point is not stationary. Positive y″ describes convexity; by itself it does not make a minimum.
12 / Read the two derivative signs separately
Use intervals and the original domain.
- Differentiate the first derivative to get the second.
- f′ describes rising or falling; f″ describes change of slope.
- State “concave up” or “concave down” to avoid terminology ambiguity.
- Use a sign chart, including exclusions and possible zero values.
- A zero second derivative alone does not prove a concavity change.
- For parametric second derivatives, divide by dx/dt again.
16 · A complete shape statement
For y = −ln x, x > 0, describe its monotonicity and concavity.
Hint
Find y′ and y″ separately.
Worked solution
y′ = −1/x < 0 and y″ = 1/x² > 0. The curve decreases and is convex throughout its real domain.
Section 1 of 12 · Differentiate the gradient