AP Calculus AB · Lesson 3

Derivatives of trig functions

Know Know it

With xx in radians, ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x and ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x. These follow from the limit definition and two special limits, lim⁡h→0sin⁡hh=1\lim_{h \to 0}\frac{\sin h}{h} = 1 and lim⁡h→0cos⁡h−1h=0\lim_{h \to 0}\frac{\cos h - 1}{h} = 0. The first is true only when hh is in radians.

The other four follow from the quotient rule. A quick memory aid: the three "co-" functions (cosine, cotangent, cosecant) have derivatives that start with a minus sign.

The six trig derivatives (xx in radians)
FunctionDerivative
sin⁡x\sin xcos⁡x\cos x
cos⁡x\cos x−sin⁡x-\sin x
tan⁡x\tan xsec⁡2x\sec^2 x
cot⁡x\cot x−csc⁡2x-\csc^2 x
sec⁡x\sec xsec⁡xtan⁡x\sec x\tan x
csc⁡x\csc x−csc⁡xcot⁡x-\csc x\cot x

Worked example Easy

What is the derivative of y=3sin⁡x−2cos⁡xy = 3\sin x - 2\cos x?

  1. 3cos⁡x−2sin⁡x3\cos x - 2\sin x
  2. 3cos⁡x+2sin⁡x3\cos x + 2\sin xAnswer
  3. −3cos⁡x+2sin⁡x-3\cos x + 2\sin x
  4. 3cos⁡x+2cos⁡x3\cos x + 2\cos x

How to solve it

  1. ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x, so 3sin⁡x→3cos⁡x3\sin x \to 3\cos x.
  2. ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x, so −2cos⁡x→2sin⁡x-2\cos x \to 2\sin x.
  3. Together: y′=3cos⁡x+2sin⁡xy' = 3\cos x + 2\sin x.

Why each choice is right or wrong

  • A. Incorrect. 3cos⁡x−2sin⁡x3\cos x - 2\sin x uses sin⁡x\sin x as the derivative of cos⁡x\cos x; the derivative is −sin⁡x-\sin x, so −2cos⁡x-2\cos x becomes +2sin⁡x+2\sin x.
  • B. Correct. ddx(3sin⁡x)=3cos⁡x\frac{d}{dx}(3\sin x) = 3\cos x and ddx(−2cos⁡x)=−2(−sin⁡x)=2sin⁡x\frac{d}{dx}(-2\cos x) = -2(-\sin x) = 2\sin x.
  • C. Incorrect. −3cos⁡x+2sin⁡x-3\cos x + 2\sin x puts a minus sign on the derivative of sin⁡x\sin x, which is +cos⁡x+\cos x.
  • D. Incorrect. 3cos⁡x+2cos⁡x3\cos x + 2\cos x uses −cos⁡x-\cos x as the derivative of cos⁡x\cos x; the sign is right but the function is wrong, since it is −sin⁡x-\sin x.

Worked example Medium

What is the slope of the graph of y=x2cos⁡xy = x^2\cos x at x=πx = \pi?

  1. −2π-2\piAnswer
  2. 2π2\pi
  3. −π2-\pi^2
  4. π2−2π\pi^2 - 2\pi

How to solve it

  1. Product rule with x2x^2 and cos⁡x\cos x: y′=2xcos⁡x+x2(−sin⁡x)y' = 2x\cos x + x^2(-\sin x).
  2. At x=πx = \pi: cos⁡π=−1\cos\pi = -1 and sin⁡π=0\sin\pi = 0.
  3. So y′=2π(−1)−π2(0)=−2πy' = 2\pi(-1) - \pi^2(0) = -2\pi.

Why each choice is right or wrong

  • A. Correct. By the product rule, y′=2xcos⁡x−x2sin⁡xy' = 2x\cos x - x^2\sin x. At π\pi, cos⁡π=−1\cos\pi = -1 and sin⁡π=0\sin\pi = 0, so y′=2π(−1)−0=−2πy' = 2\pi(-1) - 0 = -2\pi.
  • B. Incorrect. 2π2\pi comes from using cos⁡π=1\cos\pi = 1 in the first term of the product rule, but cos⁡π=−1\cos\pi = -1.
  • C. Incorrect. −π2-\pi^2 is the value of yy at π\pi, π2cos⁡π\pi^2\cos\pi, not the slope.
  • D. Incorrect. π2−2π\pi^2 - 2\pi treats sin⁡π\sin\pi as −1-1; in fact sin⁡π=0\sin\pi = 0, so the second term of the product rule vanishes.

Explain Explain it

Task verbs: explain, calculate

To see where ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^2 x comes from, write tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} and use the quotient rule: cos⁡x⋅cos⁡x−sin⁡x⋅(−sin⁡x)cos⁡2x=cos⁡2x+sin⁡2xcos⁡2x=1cos⁡2x=sec⁡2x\frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.

Trap Radians only

ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x is true only when xx is in radians. In degrees an extra factor of π180\frac{\pi}{180} appears. Calculus questions use radians unless they say otherwise, so set a calculator to radian mode.

Worked example Hard

If f(x)=sin⁡x1+cos⁡xf(x) = \frac{\sin x}{1 + \cos x}, what is f′(π3)f'\left(\frac{\pi}{3}\right)?

  1. −33-\frac{\sqrt{3}}{3}
  2. 00
  3. 23\frac{2}{3}Answer
  4. 22

How to solve it

  1. Top sin⁡x\sin x has derivative cos⁡x\cos x; bottom 1+cos⁡x1 + \cos x has derivative −sin⁡x-\sin x.
  2. Quotient rule: cos⁡x(1+cos⁡x)−sin⁡x(−sin⁡x)(1+cos⁡x)2=cos⁡x+cos⁡2x+sin⁡2x(1+cos⁡x)2\frac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2}.
  3. Since cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, this is 11+cos⁡x\frac{1}{1 + \cos x}. At π3\frac{\pi}{3}, cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}, so f′=23f' = \frac{2}{3}.

Why each choice is right or wrong

  • A. Incorrect. −33-\frac{\sqrt{3}}{3} divides the derivative of the top, cos⁡x\cos x, by the derivative of the bottom, −sin⁡x-\sin x; the quotient rule does not work that way.
  • B. Incorrect. 0 comes from subtracting sin⁡2x\sin^2 x on top. The second product is sin⁡x⋅(−sin⁡x)\sin x \cdot (-\sin x), and subtracting it adds sin⁡2x\sin^2 x.
  • C. Correct. The quotient rule gives cos⁡x(1+cos⁡x)+sin⁡2x(1+cos⁡x)2=cos⁡x+1(1+cos⁡x)2=11+cos⁡x\frac{\cos x(1 + \cos x) + \sin^2 x}{(1 + \cos x)^2} = \frac{\cos x + 1}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}, which is 11.5=23\frac{1}{1.5} = \frac{2}{3} at π3\frac{\pi}{3}.
  • D. Incorrect. 2 is sec⁡π3=1cos⁡x\sec\frac{\pi}{3} = \frac{1}{\cos x} at π3\frac{\pi}{3}, which drops the 1 from the simplified derivative 11+cos⁡x\frac{1}{1 + \cos x}.

Explain Your turn

Write sec⁡x\sec x as 1cos⁡x\frac{1}{\cos x} and use the quotient rule to find its derivative. Then show that your answer equals sec⁡xtan⁡x\sec x\tan x.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Applies the quotient rule correctly to get sin⁡xcos⁡2x\frac{\sin x}{\cos^2 x}
  • Rewrites sin⁡xcos⁡2x\frac{\sin x}{\cos^2 x} as sec⁡xtan⁡x\sec x\tan x

Model answer

With top 1 and bottom cos⁡x\cos x, the quotient rule gives 0⋅cos⁡x−1⋅(−sin⁡x)cos⁡2x=sin⁡xcos⁡2x\frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x}. Splitting the fraction, sin⁡xcos⁡2x=1cos⁡x⋅sin⁡xcos⁡x=sec⁡xtan⁡x\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x\tan x.

Apply Apply it

Calculator On the exam

On the exam a graphing calculator is required in multiple-choice Part B and free-response Part A, and it is not allowed in multiple-choice Part A or free-response Part B. In the calculator parts it can give the derivative of a function at a point as a decimal. Set it to radian mode for any trig work. Learn every rule here by hand, because many derivative questions are in the no-calculator parts.

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