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Exercise 5.5 · Q17

Q.Differentiate (x2−5x+8)(x3+7x+9)(x^2 - 5x + 8)(x^3 + 7x + 9) in three ways mentioned below:

(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation. Do they all give the same answer?
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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The derivative of (x2−5x+8)(x3+7x+9)(x^2 - 5x + 8)(x^3 + 7x + 9) is 5x4−20x3+45x2−52x+115x^4 - 20x^3 + 45x^2 - 52x + 11, and all three methods — product rule, expansion, and logarithmic differentiation — yield the same result, confirming consistency.

We are differentiating a product of two polynomials. The core idea is that the derivative of a product u⋅vu \cdot v is not simply u′⋅v′u' \cdot v' — that would be a common mistake. Instead, the product rule tells us: each piece gets a turn to be differentiated while the other stays unchanged, and we add the results. This is the heart of why the rule works: it accounts for how small changes in both factors contribute to the overall change.

Let’s work through each method step by step.


1. Using the product rule

Let u=x2−5x+8u = x^2 - 5x + 8 and v=x3+7x+9v = x^3 + 7x + 9.

First, find the derivatives:

  • u′=2x−5u' = 2x - 5
  • v′=3x2+7v' = 3x^2 + 7

The product rule states: ddx(uv)=u′v+uv′\frac{d}{dx}(u v) = u' v + u v'.

So:

dydx=(2x−5)(x3+7x+9)+(x2−5x+8)(3x2+7)\frac{dy}{dx} = (2x - 5)(x^3 + 7x + 9) + (x^2 - 5x + 8)(3x^2 + 7)

Now expand each term carefully.

First term: (2x−5)(x3+7x+9)(2x - 5)(x^3 + 7x + 9)

  • 2x⋅x3=2x42x \cdot x^3 = 2x^4
  • 2x⋅7x=14x22x \cdot 7x = 14x^2
  • 2x⋅9=18x2x \cdot 9 = 18x
  • −5⋅x3=−5x3-5 \cdot x^3 = -5x^3
  • −5⋅7x=−35x-5 \cdot 7x = -35x
  • −5⋅9=−45-5 \cdot 9 = -45

So first term = 2x4−5x3+14x2+(18x−35x)−45=2x4−5x3+14x2−17x−452x^4 - 5x^3 + 14x^2 + (18x - 35x) - 45 = 2x^4 - 5x^3 + 14x^2 - 17x - 45

Second term: (x2−5x+8)(3x2+7)(x^2 - 5x + 8)(3x^2 + 7)

  • x2⋅3x2=3x4x^2 \cdot 3x^2 = 3x^4
  • x2⋅7=7x2x^2 \cdot 7 = 7x^2
  • −5x⋅3x2=−15x3-5x \cdot 3x^2 = -15x^3
  • −5x⋅7=−35x-5x \cdot 7 = -35x
  • 8⋅3x2=24x28 \cdot 3x^2 = 24x^2
  • 8⋅7=568 \cdot 7 = 56

So second term = 3x4−15x3+(7x2+24x2)−35x+56=3x4−15x3+31x2−35x+563x^4 - 15x^3 + (7x^2 + 24x^2) - 35x + 56 = 3x^4 - 15x^3 + 31x^2 - 35x + 56

Now add them:

  • x4x^4 terms: 2x4+3x4=5x42x^4 + 3x^4 = 5x^4
  • x3x^3 terms: −5x3−15x3=−20x3-5x^3 - 15x^3 = -20x^3
  • x2x^2 terms: 14x2+31x2=45x214x^2 + 31x^2 = 45x^2
  • xx terms: −17x−35x=−52x-17x - 35x = -52x
  • Constant: −45+56=11-45 + 56 = 11

Thus:

dydx=5x4−20x3+45x2−52x+11\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11

Watch out

A common slip is forgetting to distribute the minus sign when expanding terms like −5x⋅7x-5x \cdot 7x — always double-check signs.


2. By expanding the product first

Multiply the two polynomials directly:

(x2−5x+8)(x3+7x+9)(x^2 - 5x + 8)(x^3 + 7x + 9)

Multiply each term of the first by each term of the second:

  • x2⋅x3=x5x^2 \cdot x^3 = x^5
  • x2⋅7x=7x3x^2 \cdot 7x = 7x^3
  • x2⋅9=9x2x^2 \cdot 9 = 9x^2
  • −5x⋅x3=−5x4-5x \cdot x^3 = -5x^4
  • −5x⋅7x=−35x2-5x \cdot 7x = -35x^2
  • −5x⋅9=−45x-5x \cdot 9 = -45x
  • 8⋅x3=8x38 \cdot x^3 = 8x^3
  • 8⋅7x=56x8 \cdot 7x = 56x
  • 8⋅9=728 \cdot 9 = 72

Now combine like terms:

  • x5x^5: 1x51x^5
  • x4x^4: −5x4-5x^4
  • x3x^3: 7x3+8x3=15x37x^3 + 8x^3 = 15x^3
  • x2x^2: 9x2−35x2=−26x29x^2 - 35x^2 = -26x^2
  • xx: −45x+56x=11x-45x + 56x = 11x
  • Constant: 7272

So the expanded polynomial is:

y=x5−5x4+15x3−26x2+11x+72y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72

Now differentiate term by term:

  • ddx(x5)=5x4\frac{d}{dx}(x^5) = 5x^4
  • ddx(−5x4)=−20x3\frac{d}{dx}(-5x^4) = -20x^3
  • ddx(15x3)=45x2\frac{d}{dx}(15x^3) = 45x^2
  • ddx(−26x2)=−52x\frac{d}{dx}(-26x^2) = -52x
  • ddx(11x)=11\frac{d}{dx}(11x) = 11
  • ddx(72)=0\frac{d}{dx}(72) = 0

So:

dydx=5x4−20x3+45x2−52x+11\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11

This matches exactly.

Tip

Expanding first is often easier for simple polynomials, but the product rule is essential when factors are not easily multiplied (e.g., trigonometric or logarithmic functions).


3. By logarithmic differentiation …

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