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Chapter 5 · Continuity and Differentiability

Miscellaneous · Q2: Differentiate sin⁡3x+cos⁡6x\sin^3x+\cos^6x

Miscellaneous ExerciseChain rule · Miscellaneous Exercise on Chapter 5
The question

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

2. sin⁡3x+cos⁡6x\sin^3x + \cos^6x

Exam answer

Let y=sin⁡3x+cos⁡6xy = \sin^3x + \cos^6x⁠. By the chain rule,

dydx=3sin⁡2x⋅cos⁡x+6cos⁡5x⋅(−sin⁡x)=3sin⁡2xcos⁡x−6cos⁡5xsin⁡x\begin{aligned} \frac{dy}{dx} &= 3\sin^2x\cdot\cos x \\ &\quad + 6\cos^5x\cdot(-\sin x) \\ &= 3\sin^2x\cos x - 6\cos^5x\sin x \end{aligned}

∴ dydx=3sin⁡xcos⁡x (sin⁡x−2cos⁡4x)\therefore\ \dfrac{dy}{dx} = 3\sin x\cos x\,(\sin x - 2\cos^4x)

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

The one idea this problem is built on

sin⁡3x\sin^3x means (sin⁡x)3(\sin x)^3 and cos⁡6x\cos^6x means (cos⁡x)6(\cos x)^6⁠. Each is a power of a trigonometric function, so each is differentiated by the chain rule: outer function a power, inner function sin⁡x\sin x or cos⁡x\cos x⁠. Then we take out the common factor 3sin⁡xcos⁡x3\sin x\cos x⁠.

Given, to find, and the plan
Given
y=sin⁡3x+cos⁡6xy = \sin^3x + \cos^6x⁠, defined and differentiable for every real xx⁠.
To find
dydx\dfrac{dy}{dx}⁠.
Plan
(1) Sum rule: differentiate the two terms separately. (2) Chain rule on (sin⁡x)3(\sin x)^3⁠. (3) Chain rule on (cos⁡x)6(\cos x)^6⁠. (4) Add and factorise to reach the key's form.

Basics you need

1Notation sin⁡nx\sin^n x

For a positive integer nn⁠, sin⁡nx=(sin⁡x)n\sin^n x = (\sin x)^n⁠. (This is different from sin⁡−1x\sin^{-1}x⁠, which is the inverse function.) Basics: trigonometric identities

2Chain rule for a power

ddxun=nun−1 dudx\frac{d}{dx}u^n = nu^{n-1}\,\frac{du}{dx}⁠, from the power rule (Table 5.3) and the chain rule (Theorem 4, Section 5.3.1). Lesson: chain rule

3Derivatives of sin⁡x\sin x and cos⁡x\cos x

ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠, ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x (Table 5.3). Basics: derivative rules

Solution

  1. Split with the sum rule

    Let y=(sin⁡x)3+(cos⁡x)6y = (\sin x)^3 + (\cos x)^6⁠. By the sum rule,

    dydx=ddx(sin⁡x)3+ddx(cos⁡x)6.\frac{dy}{dx} = \frac{d}{dx}(\sin x)^3 + \frac{d}{dx}(\cos x)^6.
    Writing the powers with brackets makes the structure clear: each term is a power of a trigonometric function.
    Section 5.3, rule (1). Basics: derivative rules

  2. First term: (sin⁡x)3(\sin x)^3

    Outer function u3u^3⁠, inner function u=sin⁡xu = \sin x⁠:

    ddx(sin⁡x)3=3(sin⁡x)2⋅cos⁡x=3sin⁡2xcos⁡x.\begin{aligned} \frac{d}{dx}(\sin x)^3 &= 3(\sin x)^2\cdot\cos x \\ &= 3\sin^2x\cos x. \end{aligned}
    Bring down the 3, lower the power to 2, keep sin⁡x\sin x inside, then multiply by the derivative of sin⁡x\sin x⁠, which is cos⁡x\cos x⁠.
    Basics 2 and 3. Lesson: chain rule

  3. Second term: (cos⁡x)6(\cos x)^6

    Outer function u6u^6⁠, inner function u=cos⁡xu = \cos x⁠:

    ddx(cos⁡x)6=6(cos⁡x)5⋅(−sin⁡x)=−6cos⁡5xsin⁡x.\begin{aligned} \frac{d}{dx}(\cos x)^6 &= 6(\cos x)^5\cdot(-\sin x) \\ &= -6\cos^5x\sin x. \end{aligned}
    Same pattern; the derivative of the inside, cos⁡x\cos x⁠, is −sin⁡x-\sin x⁠, which makes this term negative.
    Basics 2 and 3.

  4. Add and factorise

    dydx=3sin⁡2xcos⁡x−6cos⁡5xsin⁡x.\frac{dy}{dx} = 3\sin^2x\cos x - 6\cos^5x\sin x.

    Both terms contain 3sin⁡xcos⁡x3\sin x\cos x⁠: the first is 3sin⁡xcos⁡x⋅sin⁡x3\sin x\cos x\cdot\sin x⁠, the second is 3sin⁡xcos⁡x⋅2cos⁡4x3\sin x\cos x\cdot 2\cos^4x⁠. So

    dydx=3sin⁡xcos⁡x (sin⁡x−2cos⁡4x).\frac{dy}{dx} = 3\sin x\cos x\,(\sin x - 2\cos^4x).

    This holds for every real xx⁠.

    Taking out the common factor gives the compact form of the answer key. Multiplying back out returns the two terms, which is a quick check of the factorisation.
Result dydx=3sin⁡2xcos⁡x−6cos⁡5xsin⁡x=3sin⁡xcos⁡x (sin⁡x−2cos⁡4x),\begin{aligned} \frac{dy}{dx} &= 3\sin^2x\cos x - 6\cos^5x\sin x \\ &= 3\sin x\cos x\,(\sin x - 2\cos^4x), \end{aligned}

for every real xx⁠. (Textbook answer: 3sin⁡xcos⁡x(sin⁡x−2cos⁡4x)3\sin x\cos x(\sin x - 2\cos^4x)⁠, the same.)

Check it with numbers

At x=π4x = \frac{\pi}{4}: sin⁡x=cos⁡x=12≈0.7071068\sin x = \cos x = \frac{1}{\sqrt2} \approx 0.7071068⁠, so sin⁡xcos⁡x=12\sin x\cos x = \frac12 and cos⁡4x=14\cos^4x = \frac14⁠:

dydx=32(12−12)≈0.3106602.\frac{dy}{dx} = \frac32\left(\frac{1}{\sqrt2} - \frac12\right) \approx 0.3106602.

The unfactorised form gives 322−68\frac{3}{2\sqrt2} - \frac{6}{8} ≈1.0606602−0.75=0.3106602\approx 1.0606602 - 0.75 = 0.3106602⁠, and a central difference of yy with h=0.00001h = 0.00001 gives 0.31066020.3106602⁠.

At x=1x = 1: sin⁡1≈0.8414710\sin 1 \approx 0.8414710⁠, cos⁡1≈0.5403023\cos 1 \approx 0.5403023⁠; the formula gives ≈0.9152470\approx 0.9152470⁠, and the central difference gives 0.91524700.9152470⁠.

Common mistakes
  1. Reading sin⁡3x\sin^3x as sin⁡(x3)\sin(x^3)⁠. It is (sin⁡x)3(\sin x)^3⁠; sin⁡(x3)\sin(x^3) would have derivative 3x2cos⁡(x3)3x^2\cos(x^3)⁠.
  2. Forgetting the inner derivative. ddxcos⁡6x\frac{d}{dx}\cos^6x is not 6cos⁡5x6\cos^5x⁠; it needs the factor −sin⁡x-\sin x⁠.
  3. Factorising wrongly. 6cos⁡5xsin⁡x=3sin⁡xcos⁡x⋅2cos⁡4x6\cos^5x\sin x = 3\sin x\cos x\cdot 2\cos^4x⁠, so the bracket has 2cos⁡4x2\cos^4x⁠, not 2cos⁡5x2\cos^5x⁠.
What to take away
  1. ddxsin⁡nx=nsin⁡n−1xcos⁡x\frac{d}{dx}\sin^n x = n\sin^{n-1}x\cos x and ddxcos⁡nx=−ncos⁡n−1xsin⁡x\frac{d}{dx}\cos^n x = -n\cos^{n-1}x\sin x⁠.
  2. After differentiating, look for a common factor; it often makes the answer match the key.

Every fact used, and where it comes from

FactUsed inSource
Sum ruleStep 1Section 5.3, rule (1) (Basics: derivative rules)
ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}⁠, ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠, ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin xSteps 2, 3Table 5.3, Section 5.3
Chain ruleSteps 2, 3Theorem 4, Section 5.3.1 (Lesson: chain rule)
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