NCEA Level 2DifferentiatingAS 91262

Differentiating polynomials, fractions and all

What this lesson covers

  • A polynomial is differentiated term by term, and a constant term differentiates to zero.
  • The derivative is a new function; substituting into it gives a gradient, not a y value.
Key idea

One term at a time

  • A polynomial is a sum of terms.
  • Differentiate each term separately.
  • Then add the results back up.
Worked example

Differentiate 4x³ − 5x² + 2x − 7

  • Power times coefficient.
  • Keep each sign.
  • A constant goes to 0.
  1. y = 4x³ − 5x² + 2x − 7
  2. 4x³ → 3 × 4x² = 12x²
  3. −5x² → 2 × (−5x) = −10xkeep the sign
  4. 2x → 1 × 2 = 2
  5. −7 → 0a constant
  6. dy/dx = 12x² − 10x + 2
Worked example

Fractional coefficients

  • The rule does not change.
  • Multiply, then simplify.
  1. x³/3 → 3x²/3 = x²cancel the 3
  2. −x²/2 → −2x/2 = −xcancel the 2
  3. (2/5)x⁵ → 5 × (2/5)x⁴
  4. = 2x⁴5 × 2/5 = 2
Worked example

Expand brackets first

  • No product rule here.
  • Expand, then find dy/dx.
  1. y = (x + 3)(x − 2)
  2. = x² − 2x + 3x − 6expand
  3. = x² + x − 6collect x terms
  4. dy/dx = 2x + 1term by term
Worked example

Evaluating at fractions and negatives

  • Bracket every value.
  • Gradient 0: flat there.
  1. f′(x) = 12x² − 10x + 2
  2. f′(1/2) = 12(1/2)² − 10(1/2) + 2
  3. = 12(1/4) − 5 + 2
  4. = 3 − 5 + 2 = 0flat here
  5. f′(−1) = 12(−1)² − 10(−1) + 2
  6. = 12 + 10 + 2 = 24
Worked example

The −x² trap

  • −x² means −(x²).
  • Square, then the minus.
  1. −x² at x = −3
  2. = −(−3)²bracket the value
  3. = −(9)square first
  4. = −9then the minus
  5. (−3)² = 9lost the minus
Key idea

The derivative gives gradients

  • Substitute into f′ for a gradient.
  • Substitute into f for a height.
  • Label which you have found.
Worked example

Where is the gradient 4?

  • Set dy/dx equal to 4.
  • Height from y, not dy/dx.
  1. y = x² − 6x + 1
  2. dy/dx = 2x − 6
  3. 2x − 6 = 4gradient 4
  4. 2x = 10, so x = 5
  5. y = 5² − 6(5) + 1the original y
  6. = 25 − 30 + 1 = −4
  7. the point (5, −4)
Key idea

Check your derivative

  • The power of each term drops by one.
  • The constant term has gone.
  • The number of terms usually drops by one too.
Key idea

Where the marks are

  • Achieved: differentiate and evaluate correctly.
  • Merit: several accurate steps in a row.
  • Every step written, not just the answer.
Summary

What to take away

  • Term by term, and brackets around every substitution.
Term by termpower down, power minus one
Constantsdisappear
Fractionsmultiply, then simplify
Bracketsexpand before differentiating
Substitutingbrackets around negatives
Given gradientset dy/dx equal, solve for x
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 2 needs expanding first.
  • Question 4 wants a point.
Question
1Differentiate y = 3x⁴ − 2x² + 9.
2Differentiate y = (2x − 1)(x + 4).
3f(x) = x³/6. Find f′(−2).
4Where on y = x² + 3x is the gradient −1?
Show answer

1. dy/dx = 12x³ − 4x. 2. Expand: y = 2x² + 7x − 4, so dy/dx = 4x + 7. 3. f′(x) = x²/2, so f′(−2) = 4/2 = 2. 4. 2x + 3 = −1 gives x = −2, and y = 4 − 6 = −2: the point (−2, −2).

Worked example

The answers: questions 1 and 2

  1. 1. y = 3x⁴ − 2x² + 9
  2. dy/dx = 4 × 3x³ − 2 × 2xterm by term
  3. = 12x³ − 4xthe 9 goes
  4. 2. y = (2x − 1)(x + 4)
  5. = 2x² + 8x − x − 4expand
  6. = 2x² + 7x − 4collect x terms
  7. dy/dx = 4x + 7
Worked example

The answers: questions 3 and 4

  1. 3. f′(x) = 3x²/6 = x²/2
  2. f′(−2) = (−2)²/2brackets round −2
  3. = 4/2 = 2
  4. 4. dy/dx = 2x + 3
  5. 2x + 3 = −1, so 2x = −4gradient −1
  6. x = −2
  7. y = (−2)² + 3(−2)the original y
  8. = 4 − 6 = −2
  9. the point (−2, −2)

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