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Chapter 5 · Continuity and Differentiability

Miscellaneous · Q1: Differentiate (3x2−9x+5)9(3x^2-9x+5)^9

Miscellaneous ExerciseChain rule · Miscellaneous Exercise on Chapter 5
The question

Differentiate w.r.t. xx the function in Exercises 1 to 11.

1. (3x2−9x+5)9(3x^2 - 9x + 5)^9

Exam answer

Let y=(3x2−9x+5)9y = (3x^2 - 9x + 5)^9⁠. By the chain rule,

dydx=9(3x2−9x+5)8×ddx(3x2−9x+5)=9(3x2−9x+5)8(6x−9)\begin{aligned} \frac{dy}{dx} &= 9(3x^2 - 9x + 5)^8 \\ &\quad\times\frac{d}{dx}(3x^2 - 9x + 5) \\ &= 9(3x^2 - 9x + 5)^8(6x - 9) \end{aligned}

∴ dydx=27(3x2−9x+5)8(2x−3)\therefore\ \dfrac{dy}{dx} = 27(3x^2 - 9x + 5)^8(2x - 3)

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

This is a power of a polynomial: an outer function u9u^9 applied to an inner function u=3x2−9x+5u = 3x^2 - 9x + 5⁠. The chain rule says: differentiate the outer power (keeping the inside as it is), then multiply by the derivative of the inside,

dydx=9u8⋅dudx.\frac{dy}{dx} = 9u^8\cdot\frac{du}{dx}.

Expanding the ninth power first would be correct but very long; the chain rule avoids it.

Given, to find, and the plan
Given
y=(3x2−9x+5)9y = (3x^2 - 9x + 5)^9⁠, a polynomial, defined and differentiable for every real xx⁠.
To find
dydx\dfrac{dy}{dx}⁠.
Plan
(1) Name the inner and outer functions. (2) Differentiate each. (3) Multiply (chain rule) and substitute back. (4) Take out the common factor 3 to reach the key's form.

Basics you need

1Chain rule

If y=f(u)y = f(u) and u=g(x)u = g(x)⁠, then dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} (Theorem 4, Section 5.3.1). Lesson: chain rule

2Power rule

dduun=nun−1\frac{d}{du}u^n = nu^{n-1} (Table 5.3). Basics: derivative rules

3Derivative of a polynomial

Term by term (sum rule, Section 5.3, rule (1)), with the power rule; constant multiples stay in front and the derivative of a constant is 00 (both Class XI, Limits and Derivatives).

4Composite functions

(3x2−9x+5)9(3x^2 - 9x + 5)^9 is "first compute 3x2−9x+53x^2 - 9x + 5⁠, then raise it to the 9th power". Basics: composite functions

Solution

  1. Name the inner and outer functions

    Let y=(3x2−9x+5)9y = (3x^2 - 9x + 5)^9⁠. Put

    u=3x2−9x+5,y=u9.u = 3x^2 - 9x + 5, \qquad y = u^9.
    The last operation done when computing yy is "raise to the 9th power", so that is the outer function; what it is applied to is the inner function. Naming them makes the chain rule mechanical.
    Basic 4. Basics: composite functions

  2. Differentiate each

    dydu=9u8,dudx=6x−9.\begin{aligned} \frac{dy}{du} &= 9u^8, \\ \frac{du}{dx} &= 6x - 9. \end{aligned}
    Outer: power rule with n=9n = 9⁠. Inner: term by term, ddx3x2=6x\frac{d}{dx}3x^2 = 6x⁠, ddx(−9x)=−9\frac{d}{dx}(-9x) = -9⁠, ddx5=0\frac{d}{dx}5 = 0⁠.
    Basics 2 and 3; Table 5.3. Basics: derivative rules

  3. Chain rule, then substitute back

    dydx=dydu⋅dudx=9u8(6x−9)=9(3x2−9x+5)8(6x−9).\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du}\cdot\frac{du}{dx} = 9u^8(6x - 9) \\ &= 9(3x^2 - 9x + 5)^8(6x - 9). \end{aligned}
    The answer must be in terms of xx⁠, so we replace uu by 3x2−9x+53x^2 - 9x + 5⁠. Note the inside of the bracket is unchanged; only the power dropped from 9 to 8.
    Chain rule, Theorem 4, Section 5.3.1. Lesson: chain rule

  4. Tidy up: take out the factor 3

    6x−9=3(2x−3)6x - 9 = 3(2x - 3)⁠, so

    dydx=27(3x2−9x+5)8(2x−3).\frac{dy}{dx} = 27(3x^2 - 9x + 5)^8(2x - 3).

    This holds for every real xx⁠.

    9×3=279 \times 3 = 27⁠. This matches the answer key exactly. It also shows at a glance where the derivative is zero: at x=32x = \frac32⁠, and wherever 3x2−9x+5=03x^2 - 9x + 5 = 0⁠.
Result ddx(3x2−9x+5)9=27(3x2−9x+5)8(2x−3),\begin{aligned} &\frac{d}{dx}(3x^2 - 9x + 5)^9 \\ &= 27(3x^2 - 9x + 5)^8(2x - 3), \end{aligned}

for every real xx⁠. (Textbook answer: 27(3x2−9x+5)8(2x−3)27(3x^2 - 9x + 5)^8(2x - 3)⁠, the same.)

Check it with numbers

At x=1x = 1: the inside is 3−9+5=−13 - 9 + 5 = -1⁠, so dydx=27⋅(−1)8⋅(2−3)=−27\frac{dy}{dx} = 27\cdot(-1)^8\cdot(2 - 3) = -27⁠. A central difference of yy with h=0.00001h = 0.00001 gives −27.00000-27.00000⁠.

At x=2x = 2: the inside is 12−18+5=−112 - 18 + 5 = -1 again, so dydx=27⋅1⋅(4−3)=27\frac{dy}{dx} = 27\cdot 1\cdot(4 - 3) = 27⁠; the central difference gives 27.0000027.00000⁠.

Common mistakes
  1. Forgetting the inner derivative. 9(3x2−9x+5)89(3x^2 - 9x + 5)^8 alone is wrong; it must be multiplied by 6x−96x - 9⁠.
  2. Differentiating the inside inside the bracket. The bracket stays as 3x2−9x+53x^2 - 9x + 5⁠; writing (6x−9)8(6x - 9)^8 is a common slip.
  3. Arithmetic in the tidy-up. 9(6x−9)=27(2x−3)9(6x - 9) = 27(2x - 3)⁠, not 27(2x−9)27(2x - 9)⁠.
What to take away
  1. ddx[g(x)]n=n[g(x)]n−1 g′(x)\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1}\,g'(x)⁠: bring down the power, lower it by one, multiply by the derivative of the inside.
  2. Factor out common numbers at the end to match standard answers.

Every fact used, and where it comes from

FactUsed inSource
ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}Step 2Table 5.3, Section 5.3 (Basics: derivative rules)
Sum ruleStep 2Section 5.3, rule (1)
Constant multiples; the derivative of a constant is 0Step 2Class XI, Limits and Derivatives
Chain ruleStep 3Theorem 4, Section 5.3.1 (Lesson: chain rule)
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