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B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Finding the Second Derivative of Polynomial and Rational Functions

B1.3About 65 minutes
Learning goals
  • I can find f″(x) for polynomial functions using the power, sum and chain rules.
  • I can find f″(x) for simple rational functions by rewriting with negative exponents or using the quotient rule.
  • I can simplify and factor f″(x) so its sign is easy to read.

Words to know

negative exponent
x⁻ⁿ = 1/xⁿ. Rewriting fractions this way lets you use the power rule.
quotient rule
(u/v)′ = (u′v − uv′)/v², which you derived from the product and chain rules in Derivatives and Differentiation Rules.
simple rational function
A quotient of two polynomials of low degree, such as 1/(x − 2), x/(x + 1) or 1/(x² + 1).

Two derivatives in a row

In the last lesson you met f″ through graphs and data. Now you will find it algebraically, quickly and accurately, for polynomial and simple rational functions. Every later lesson in this unit depends on this skill, so it is worth practising until it is automatic.

The plan is simple: find f′, simplify it, then differentiate again. A tidy f′ makes f″ much easier. Here are the rules you will use (from Derivatives and Differentiation Rules):

  • Power rule: d/dx xⁿ = nxⁿ⁻¹ for any rational n (including negative n).
  • Constant multiple, sum and difference rules.
  • Chain rule: d/dx [g(x)]ⁿ = n[g(x)]ⁿ⁻¹ g′(x).
  • Product rule: (uv)′ = u′v + uv′.
Worked example

Example 1: A polynomial

Find f″(x) for f(x) = x⁴ − 2x³ + 5x − 8, and evaluate f″(−1) and f″(2).

  1. f′(x) = 4x³ − 6x² + 5.
  2. f″(x) = 12x² − 12x = 12x(x − 1).
  3. f″(−1) = 12(1) + 12 = 24 > 0, so f is concave up at x = −1.
  4. f″(2) = 48 − 24 = 24 > 0. f″ = 0 at x = 0 and x = 1, where the concavity may change.

f″(x) = 12x² − 12x; f″(−1) = 24 and f″(2) = 24.

Worked example

Example 2: Chain rule twice

Find f″(x) for f(x) = (2x − 1)⁵.

  1. f′(x) = 5(2x − 1)⁴ · 2 = 10(2x − 1)⁴.
  2. f″(x) = 10 · 4(2x − 1)³ · 2 = 80(2x − 1)³.

f″(x) = 80(2x − 1)³. (No need to expand: the factored form shows f″ = 0 at x = 1/2 and its sign changes there.)

Rational functions: rewrite with negative exponents

For a simple rational function, rewrite before differentiating:

  • 1/x = x⁻¹, and 3/(x − 2)² = 3(x − 2)⁻².
  • If the numerator and denominator have the same degree, divide first: x/(x + 1) = 1 − 1/(x + 1).

Then use the power and chain rules. When you are finished, rewrite negative exponents as fractions so you can read signs easily.

Rewrite first, then differentiate twicef(x) = x/(x + 1)rewrite: 1 − (x + 1)⁻¹f′(x) = (x + 1)⁻²power + chain rulef″(x) = −2(x + 1)⁻³= −2/(x + 1)³
Rewriting a rational function with negative exponents turns each derivative into a power rule and chain rule step.
Worked example

Example 3: Simple rational functions

Find f″(x) for (a) f(x) = 1/x, (b) g(x) = 3/(x − 2)², (c) h(x) = x/(x + 1).

  1. (a) f(x) = x⁻¹, f′(x) = −x⁻², f″(x) = 2x⁻³ = 2/x³.
  2. (b) g(x) = 3(x − 2)⁻², g′(x) = −6(x − 2)⁻³, g″(x) = 18(x − 2)⁻⁴ = 18/(x − 2)⁴.
  3. (c) h(x) = (x + 1 − 1)/(x + 1) = 1 − (x + 1)⁻¹. h′(x) = (x + 1)⁻². h″(x) = −2(x + 1)⁻³ = −2/(x + 1)³.

f″ = 2/x³, g″ = 18/(x − 2)⁴ (always positive, so g is concave up on both sides of x = 2), h″ = −2/(x + 1)³.

f(x) = 1/xxy−2−112−4−3−2−1123f″(x) = 2/x³xy−2−112−4−3−2−1123
For f(x) = 1/x, f″(x) = 2/x³ is negative for x < 0 (the left branch bends like ∩) and positive for x > 0 (the right branch bends like ∪). There is no inflection point, because x = 0 is not in the domain.

The quotient rule (a recap)

**Remember from Derivatives and Differentiation Rules (lesson The Quotient Rule and Rational Functions): writing u/v as u · v⁻¹ and using the product and chain rules gives the quotient rule**

(u/v)′ = (u′v − uv′)/v² ("low d-high minus high d-low, over the square of what's below").

Go back to that lesson if you want the derivation or more practice. You may use the formula or the product-rule method; both give the same answer. For a second derivative of a quotient, simplify f′ fully first, then use the rule again (or rewrite f′ with a negative exponent).

Worked example

Example 4: A bell-shaped curve

Find f′(x) and f″(x) for f(x) = 1/(x² + 1). Where is f″(x) = 0?

  1. f(x) = (x² + 1)⁻¹, so f′(x) = −(x² + 1)⁻² · 2x = −2x(x² + 1)⁻².
  2. Product rule on f′: f″(x) = −2(x² + 1)⁻² + (−2x)(−2)(x² + 1)⁻³(2x) = −2(x² + 1)⁻² + 8x²(x² + 1)⁻³.
  3. Common factor (x² + 1)⁻³: f″(x) = [−2(x² + 1) + 8x²]/(x² + 1)³ = (6x² − 2)/(x² + 1)³.
  4. f″ = 0 when 6x² = 2, x² = 1/3, x = ±1/√3 ≈ ±0.577. There, f = 1/(1/3 + 1) = 3/4.

f′(x) = −2x/(x² + 1)², f″(x) = (6x² − 2)/(x² + 1)³, which is zero at x = ±1/√3. Since the denominator is always positive, f″ changes sign there: inflection points (±1/√3, 3/4).

f(x) = 1/(x² + 1)xy−2−1120.51IPf″(x) = (6x² − 2)/(x² + 1)³xy−2−112−2−10.5(0, −2)
Where f″ < 0 (shaded, between x = −1/√3 and x = 1/√3) the bell-shaped curve is concave down; outside, f″ > 0 and it is concave up. The inflection points are at x = ±1/√3 ≈ ±0.577, where y = 3/4.

Factor before you finish

When f″ is a fraction, pull out the lowest power of each repeated factor (as in Example 4), combine into one fraction, and factor the numerator. The sign of f″ then comes from a short list of factors, exactly as with f′.

Just for fun

Why was the second derivative so calm in a crisis?

It had already been through the change of a change.

Hands-on activity

Derivative chains with cards

You need: 12 index cards (or squares of paper), a pen and a timer.

  1. On 4 cards write functions: x⁴ − 3x², (3x + 1)⁴, 2/x, x/(x − 3). On 4 more write their first derivatives, and on the last 4 their second derivatives (work them out first and check them with this lesson's examples and a partner or a grown-up).
  2. Shuffle all 12 cards. Set a timer for 3 minutes and arrange them in 4 chains: f → f′ → f″.
  3. Then turn over the f cards only and rebuild the other cards from memory.

What to look for (answers): x⁴ − 3x² → 4x³ − 6x → 12x² − 6. (3x + 1)⁴ → 12(3x + 1)³ → 108(3x + 1)². 2/x → −2/x² → 4/x³. x/(x − 3) = 1 + 3/(x − 3) → −3/(x − 3)² → 6/(x − 3)³.

Your turn

1

Find f″(x) for f(x) = 2x³ − 9x² + 12x.

2

Find f″(x) for f(x) = (x² + 3)².

3

Find y″ for y = 4/x².

4

Find f″(x) for f(x) = 1/(x − 5).

5

Find f″(x) for f(x) = (x + 2)/(x − 1).

6

For f(x) = x⁴ − 6x², evaluate f″(0) and f″(2) and state the concavity at each.

7

Find where f″(x) = 0 for f(x) = x⁵ − 5x⁴.

More practice: warm-up

1

Find f″(x) for f(x) = 5x³.

2

Find f″(x) for f(x) = x² − 7x + 1.

3

Rewrite 3/x⁴ with a negative exponent.

4

Find f″(x) for f(x) = x⁻¹.

5

Find f″(1) for f(x) = x⁶.

More practice: on your own

1

Find f″(x) for f(x) = (3x − 2)⁴.

2

Find f″(x) for f(x) = x²/(x + 1). (Hint: divide first.)

3

Find f″(x) for f(x) = √x (x > 0).

4

Use the quotient rule to find f′(x) for f(x) = x/(x² + 4).

5

Find f″(x) for f(x) = x(x − 3)².

Challenge

1

Find f″(x) for f(x) = 2x/(x² + 1) and simplify fully.

2

Find a formula for the nth derivative of f(x) = 1/x.

3

Find all values of k for which f(x) = x⁴ + kx³ + 6x² is concave up for every x.

Check your understanding

Quiz

1.

What is f″(x) for f(x) = x³ − 4x² + x?

2.

What is f″(x) for f(x) = 1/x²?

3.

What is f″(x) for f(x) = (x − 3)⁴?

4.

Which rewriting is the best first step to differentiate f(x) = x/(x − 2)?

5.

For f(x) = 1/(x² + 1), what is f″(0)?

6.

The quotient rule says (u/v)′ equals:

For parents and teachers

Big idea (B1.3): finding f″ is two derivatives in a row. For rational functions, rewriting with negative exponents (and dividing first when the degrees are equal) keeps the algebra short.

How to help: ask your teen to check one answer by a second method (for example, expand (x² + 3)² and differentiate, or use the quotient rule instead of negative exponents). Matching answers build confidence.

Common difficulties:

  • Forgetting the chain-rule factor (the derivative of the inside).
  • Sign errors with negative exponents: the derivative of x⁻¹ is −x⁻², and of −x⁻² is +2x⁻³.
  • Not simplifying f′ before differentiating again.
  • Writing the quotient-rule numerator in the wrong order.