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The Derivative & Its Rules · Topic 08 of 23

Derivatives of Trigonometric Functions

In radians, (sin⁡x)′=cos⁡x(\sin x)'=\cos x and (cos⁡x)′=−sin⁡x(\cos x)'=-\sin x; the quotient rule then gives (tan⁡x)′=sec⁡2x(\tan x)'=\sec^2x, (cot⁡x)′=−csc⁡2x(\cot x)'=-\csc^2x, (sec⁡x)′=sec⁡xtan⁡x(\sec x)'=\sec x\tan x and (csc⁡x)′=−csc⁡xcot⁡x(\csc x)'=-\csc x\cot x. The three co-functions carry the minus sign, each formula holds only where the function is defined, and anything but a bare xx inside needs the chain rule.

4 min readCore — on every final8 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Sine and cosine trade places

    (sin⁡x)′=cos⁡x(\sin x)'=\cos x and (cos⁡x)′=−sin⁡x(\cos x)'=-\sin x. Both come from the definition: expand sin⁡(x+h)\sin(x+h) with the addition formula, then use sin⁡hh→1\frac{\sin h}{h}\to1 and cos⁡h−1h→0\frac{\cos h-1}{h}\to0. Where sin⁡\sin peaks, its slope is cos⁡x=0\cos x=0.

  2. 2

    Co-functions carry the minus sign

    Co-functions cos⁡\cos, cot⁡\cot and csc⁡\csc carry the minus sign in their derivatives; sin⁡\sin, tan⁡\tan and sec⁡\sec do not. Swap each function for its co-function and flip the sign to turn one row of the table into its partner.

  3. 3

    Quotient rule builds the other four

    Write tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x} and sec⁡x=1cos⁡x\sec x=\frac{1}{\cos x}, apply the quotient rule, and sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 finishes the algebra. Each result holds exactly where its denominator is nonzero — never at a vertical asymptote.

  4. 4

    Radians, or nothing works

    Every formula rests on lim⁡x→0sin⁡xx=1\lim_{x\to0}\frac{\sin x}{x}=1, which is only true in radians. In degrees each differentiation brings in a factor π180\frac{\pi}{180}. Keep the calculator in radians whenever a derivative is involved.

  5. 5

    Higher derivatives cycle in fours

    sin⁡x→cos⁡x→−sin⁡x→−cos⁡x→sin⁡x\sin x\to\cos x\to-\sin x\to-\cos x\to\sin x, so the nnth derivative depends only on the remainder of nn on division by 44. In particular y=sin⁡xy=\sin x and y=cos⁡xy=\cos x both satisfy y′′+y=0y''+y=0.

π/2π3π/22π-1.0-0.50.51.0xytangent, slope 0cos(π/2) = 0cos(3π/2) = 0f(x) = sin xf′(x) = cos x
f(x)=sin⁡xf(x)=\sin x (blue) with its derivative f′(x)=cos⁡xf'(x)=\cos x (amber, dashed). Where sin⁡\sin has a maximum or minimum the tangent is flat, and the derivative crosses zero at exactly those xx: cos⁡π2=cos⁡3π2=0\cos\frac{\pi}{2}=\cos\frac{3\pi}{2}=0. Between them sin⁡\sin decreases and cos⁡\cos is negative.
OAPT1θsin θtan θareas: ½ sin θ ≤ ½ θ ≤ ½ tan θso cos θ ≤ sin θ / θ ≤ 1
The squeeze behind lim⁡θ→0sin⁡θθ=1\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1, with P=(cos⁡θ,sin⁡θ)P=(\cos\theta,\sin\theta) on the unit circle. Triangle OAPOAP (area 12sin⁡θ\frac12\sin\theta) fits inside the shaded sector (area 12θ\frac12\theta), which fits inside triangle OATOAT (area 12tan⁡θ\frac12\tan\theta); divide by 12sin⁡θ\frac12\sin\theta and invert to get cos⁡θ≤sin⁡θθ≤1\cos\theta\le\frac{\sin\theta}{\theta}\le1. The sector area is 12θ\frac12\theta only in radians — that is where the whole table gets its unit.
02

Formulas

What to have memorised

  • Sine and cosine

    ddxsin⁡x=cos⁡xddxcos⁡x=−sin⁡x\frac{d}{dx}\sin x=\cos x\qquad \frac{d}{dx}\cos x=-\sin x

    All real xx, measured in radians.

  • Tangent and cotangent

    ddxtan⁡x=sec⁡2xddxcot⁡x=−csc⁡2x\frac{d}{dx}\tan x=\sec^2 x\qquad \frac{d}{dx}\cot x=-\csc^2 x

    Valid for x≠π2+kπx\neq\frac{\pi}{2}+k\pi and x≠kπx\neq k\pi respectively.

  • Secant and cosecant

    ddxsec⁡x=sec⁡xtan⁡xddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx}\sec x=\sec x\tan x\qquad \frac{d}{dx}\csc x=-\csc x\cot x

    Each derivative starts with the function itself; only csc⁡\csc takes the minus.

  • The two limits behind the table

    lim⁡h→0sin⁡hh=1lim⁡h→0cos⁡h−1h=0\lim_{h\to0}\frac{\sin h}{h}=1\qquad \lim_{h\to0}\frac{\cos h-1}{h}=0

    Radians only. Prove the first by squeezing, not by l'Hôpital — that is circular.

  • Angle addition

    sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h)=\sin x\cos h+\cos x\sin h

    For the definition proof; cos⁡(x+h)=cos⁡xcos⁡h−sin⁡xsin⁡h\cos(x+h)=\cos x\cos h-\sin x\sin h.

  • Identities that finish the algebra

    sin⁡2x+cos⁡2x=11+tan⁡2x=sec⁡2x1+cot⁡2x=csc⁡2x\sin^2x+\cos^2x=1\qquad 1+\tan^2x=\sec^2x\qquad 1+\cot^2x=\csc^2x

    Divide the first by cos⁡2x\cos^2x or sin⁡2x\sin^2x to get the other two.

  • nnth derivative of sine

    dndxnsin⁡x=sin⁡(x+nπ2)\frac{d^n}{dx^n}\sin x=\sin\left(x+\frac{n\pi}{2}\right)

    Period four: n=1,2,3,4n=1,2,3,4 gives cos⁡x, −sin⁡x, −cos⁡x, sin⁡x\cos x,\,-\sin x,\,-\cos x,\,\sin x.

  • Inside function? Chain rule

    ddxsin⁡(u(x))=cos⁡(u(x)) u′(x)\frac{d}{dx}\sin\bigl(u(x)\bigr)=\cos\bigl(u(x)\bigr)\,u'(x)

    The table is for a bare xx; sin⁡3x\sin 3x and sin⁡(x2)\sin(x^2) need the inside derivative.

03

Differentiate a trig expression

The steps, in order
  1. 1

    Look at the argument first: a bare xx means the table applies directly, while sin⁡3x\sin 3x or cos⁡(x2)\cos(x^2) also needs the chain rule.

  2. 2

    Name the structure — sum, product or quotient — and write uu, vv, u′u', v′v' before combining anything.

  3. 3

    Apply the rule with its signs: u′v+uv′u'v+uv' for a product, u′v−uv′v2\frac{u'v-uv'}{v^2} for a quotient, and −sin⁡x-\sin x, −csc⁡2x-\csc^2x, −csc⁡xcot⁡x-\csc x\cot x for the co-functions.

  4. 4

    Simplify with sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 or 1+tan⁡2x=sec⁡2x1+\tan^2x=\sec^2x and cancel common factors — most exam answers collapse.

  5. 5

    For a tangent line, evaluate f(a)f(a) and f′(a)f'(a) with exact values (sin⁡π3=32\sin\frac{\pi}{3}=\frac{\sqrt3}{2}, not 0.8660.866), then write y=f(a)+f′(a)(x−a)y=f(a)+f'(a)(x-a).

  6. 6

    State validity by excluding every xx that makes a denominator zero, including the ones hidden inside tan⁡\tan, sec⁡\sec, cot⁡\cot and csc⁡\csc.

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    ddxcos⁡x=sin⁡x\frac{d}{dx}\cos x=\sin x

    ✓ Right

    ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x=-\sin x

    Why: The three co-functions (cos⁡\cos, cot⁡\cot, csc⁡\csc) carry the minus sign.

  • ✗ Wrong

    ddxsec⁡x=−sec⁡xtan⁡x\frac{d}{dx}\sec x=-\sec x\tan x

    ✓ Right

    ddxsec⁡x=+sec⁡xtan⁡x\frac{d}{dx}\sec x=+\sec x\tan x; it is csc⁡\csc that gets the minus.

    Why: sec⁡\sec is not a co-function.

  • ✗ Wrong

    ddx(sin⁡xcos⁡x)=(cos⁡x)(−sin⁡x)\frac{d}{dx}(\sin x\cos x)=(\cos x)(-\sin x)

    ✓ Right

    Product rule: cos⁡x⋅cos⁡x+sin⁡x⋅(−sin⁡x)=cos⁡2x−sin⁡2x\cos x\cdot\cos x+\sin x\cdot(-\sin x)=\cos^2x-\sin^2x.

    Why: The derivative of a product is not the product of the derivatives.

  • ✗ Wrong

    ddxsin⁡(x2)=cos⁡(x2)\frac{d}{dx}\sin(x^2)=\cos(x^2)

    ✓ Right

    2xcos⁡(x2)2x\cos(x^2) — the table is for sin⁡\sin of xx itself; anything else needs the chain rule.

  • ✗ Wrong

    (uv)′=uv′−u′vv2\left(\frac{u}{v}\right)'=\frac{uv'-u'v}{v^2}

    ✓ Right

    (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}: derivative of the top first.

    Why: Reversing the order negates the whole answer.

  • ✗ Wrong

    ddθsin⁡θ=cos⁡θ\frac{d}{d\theta}\sin\theta=\cos\theta with θ\theta in degrees

    ✓ Right

    Convert first: sin⁡(πθ180)\sin\left(\frac{\pi\theta}{180}\right) has derivative π180cos⁡(πθ180)\frac{\pi}{180}\cos\left(\frac{\pi\theta}{180}\right).

    Why: sin⁡xx→1\frac{\sin x}{x}\to1 only in radians.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Differentiate g(x)=x2cos⁡xg(x)=x^2\cos x and evaluate g′(π)g'(\pi).

    Hint

    A product of two functions of xx — the table alone is not enough.

    Show answer

    Answer

    g′(x)=2xcos⁡x−x2sin⁡xg'(x)=2x\cos x-x^2\sin x; g′(π)=−2πg'(\pi)=-2\pi.

    Steps

    Product rule with u=x2u=x^2, v=cos⁡xv=\cos x:

    g′(x)=(2x)(cos⁡x)+(x2)(−sin⁡x)=2xcos⁡x−x2sin⁡x.g'(x)=(2x)(\cos x)+(x^2)(-\sin x)=2x\cos x-x^2\sin x.

    At x=πx=\pi: cos⁡π=−1\cos\pi=-1 and sin⁡π=0\sin\pi=0, so g′(π)=2π(−1)−π2(0)=−2πg'(\pi)=2\pi(-1)-\pi^2(0)=-2\pi.

  2. Q2easy

    Starting from csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}, use the quotient rule to show that ddxcsc⁡x=−csc⁡xcot⁡x\dfrac{d}{dx}\csc x=-\csc x\cot x, and state exactly where the formula is valid.

    Hint

    Numerator 11, denominator sin⁡x\sin x; then split cos⁡xsin⁡2x\frac{\cos x}{\sin^2x} into two familiar ratios.

    Show answer

    Answer

    ddxcsc⁡x=−csc⁡xcot⁡x\dfrac{d}{dx}\csc x=-\csc x\cot x for all x≠kπx\neq k\pi, kk an integer.

    Steps

    With u=1u=1, v=sin⁡xv=\sin x: u′=0u'=0, v′=cos⁡xv'=\cos x.

    ddxcsc⁡x=(0)(sin⁡x)−(1)(cos⁡x)sin⁡2x=−cos⁡xsin⁡2x=−1sin⁡x⋅cos⁡xsin⁡x=−csc⁡xcot⁡x.\frac{d}{dx}\csc x=\frac{(0)(\sin x)-(1)(\cos x)}{\sin^2x}=-\frac{\cos x}{\sin^2x}=-\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}=-\csc x\cot x.

    The quotient rule needs v=sin⁡x≠0v=\sin x\neq0, so the formula holds exactly for x≠kπx\neq k\pi. At x=kπx=k\pi the function csc⁡x\csc x is undefined — there is no derivative to speak of.

  3. Q3medium

    Find an equation of the tangent line to y=sec⁡xy=\sec x at x=π3x=\dfrac{\pi}{3}.

    Hint

    Point first, then slope from (sec⁡x)′=sec⁡xtan⁡x(\sec x)'=\sec x\tan x, all in exact values.

    Show answer

    Answer

    y=2+23(x−π3)y=2+2\sqrt{3}\left(x-\dfrac{\pi}{3}\right).

    Steps

    Point: sec⁡π3=1cos⁡(π/3)=11/2=2\sec\frac{\pi}{3}=\dfrac{1}{\cos(\pi/3)}=\dfrac{1}{1/2}=2, so the curve passes through (π3,2)\left(\frac{\pi}{3},2\right).

    Slope: y′=sec⁡xtan⁡xy'=\sec x\tan x, so m=sec⁡π3tan⁡π3=2⋅3=23m=\sec\frac{\pi}{3}\tan\frac{\pi}{3}=2\cdot\sqrt3=2\sqrt3.

    y−2=23(x−π3)ory=23 x+2−23 π3y-2=2\sqrt3\left(x-\frac{\pi}{3}\right)\qquad\text{or}\qquad y=2\sqrt3\,x+2-\frac{2\sqrt3\,\pi}{3}

    Either form is a full-marks answer; do not round 232\sqrt3 or π3\frac{\pi}{3}.

  4. Q4medium

    Let y=sin⁡x+cos⁡xy=\sin x+\cos x on [0,2π)[0,2\pi).

    (a) Find every xx where the tangent line is horizontal, and give the corresponding points. (b) Find every xx where the tangent line has slope 11.

    Hint

    Both parts are equations in y′=cos⁡x−sin⁡xy'=\cos x-\sin x. In (b), squaring creates extra roots — check every candidate in the original equation.

    Show answer

    Answer

    (a) x=π4x=\frac{\pi}{4} and x=5π4x=\frac{5\pi}{4}, at (π4,2)\left(\frac{\pi}{4},\sqrt2\right) and (5π4,−2)\left(\frac{5\pi}{4},-\sqrt2\right). (b) x=0x=0 and x=3π2x=\frac{3\pi}{2}.

    Steps

    y′=cos⁡x−sin⁡xy'=\cos x-\sin x.

    (a) y′=0y'=0 means sin⁡x=cos⁡x\sin x=\cos x. Since cos⁡x=0\cos x=0 would force sin⁡x=0\sin x=0 too, divide: tan⁡x=1\tan x=1, so x=π4,5π4x=\frac{\pi}{4},\frac{5\pi}{4}. There y=22+22=2y=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2 and y=−22−22=−2y=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2.

    (b) cos⁡x−sin⁡x=1\cos x-\sin x=1. Squaring: 1−2sin⁡xcos⁡x=11-2\sin x\cos x=1, so sin⁡xcos⁡x=0\sin x\cos x=0, candidates 0,π2,π,3π20,\frac{\pi}{2},\pi,\frac{3\pi}{2}. Check each in the original: x=0x=0 gives 1−0=11-0=1 ✓; x=π2x=\frac{\pi}{2} gives 0−1=−10-1=-1 ✗; x=πx=\pi gives −1−0=−1-1-0=-1 ✗; x=3π2x=\frac{3\pi}{2} gives 0−(−1)=10-(-1)=1 ✓.

  5. Q5medium

    Let y=sin⁡xy=\sin x. (a) Show that y′′+y=0y''+y=0. (b) Find d2026dx2026sin⁡x\dfrac{d^{2026}}{dx^{2026}}\sin x, with justification.

    Hint

    Differentiate four times and watch what comes back.

    Show answer

    Answer

    (a) y′′=−sin⁡xy''=-\sin x, so y′′+y=0y''+y=0. (b) −sin⁡x-\sin x.

    Steps

    (a) y′=cos⁡xy'=\cos x and y′′=−sin⁡xy''=-\sin x, so y′′+y=−sin⁡x+sin⁡x=0y''+y=-\sin x+\sin x=0 for every xx.

    (b) y(1)=cos⁡xy^{(1)}=\cos x, y(2)=−sin⁡xy^{(2)}=-\sin x, y(3)=−cos⁡xy^{(3)}=-\cos x, y(4)=sin⁡xy^{(4)}=\sin x: the derivatives repeat with period 44, so only the remainder of nn on division by 44 matters. 2026=4(506)+22026=4(506)+2, remainder 22:

    d2026dx2026sin⁡x=y(2)=−sin⁡x.\frac{d^{2026}}{dx^{2026}}\sin x=y^{(2)}=-\sin x.

    Check with the closed form: sin⁡(x+2026π2)=sin⁡(x+1013π)=−sin⁡x\sin\left(x+\frac{2026\pi}{2}\right)=\sin(x+1013\pi)=-\sin x, since 10131013 is odd.

  6. Q6hard

    Differentiate f(x)=xsin⁡x1+cos⁡xf(x)=\dfrac{x\sin x}{1+\cos x} and simplify to a single fraction with no products of trig functions. State where the result is valid.

    Hint

    Product rule on the numerator first, then the quotient rule; the expanded top has a common factor of 1+cos⁡x1+\cos x.

    Show answer

    Answer

    f′(x)=x+sin⁡x1+cos⁡xf'(x)=\dfrac{x+\sin x}{1+\cos x} for x≠π+2kπx\neq\pi+2k\pi.

    Steps

    u=xsin⁡xu=x\sin x, u′=sin⁡x+xcos⁡xu'=\sin x+x\cos x; v=1+cos⁡xv=1+\cos x, v′=−sin⁡xv'=-\sin x.

    f′(x)=(sin⁡x+xcos⁡x)(1+cos⁡x)−(xsin⁡x)(−sin⁡x)(1+cos⁡x)2f'(x)=\frac{(\sin x+x\cos x)(1+\cos x)-(x\sin x)(-\sin x)}{(1+\cos x)^2}

    Numerator: sin⁡x+sin⁡xcos⁡x+xcos⁡x+xcos⁡2x+xsin⁡2x=sin⁡x(1+cos⁡x)+x(1+cos⁡x)=(1+cos⁡x)(x+sin⁡x)\sin x+\sin x\cos x+x\cos x+x\cos^2x+x\sin^2x=\sin x(1+\cos x)+x(1+\cos x)=(1+\cos x)(x+\sin x), using cos⁡2x+sin⁡2x=1\cos^2x+\sin^2x=1.

    Cancel one factor of 1+cos⁡x1+\cos x:

    f′(x)=x+sin⁡x1+cos⁡x,x≠π+2kπ.f'(x)=\frac{x+\sin x}{1+\cos x},\qquad x\neq\pi+2k\pi.

    Check at x=π2x=\frac{\pi}{2}: the formula gives 1+π21+\frac{\pi}{2}; since f(x)=xtan⁡x2f(x)=x\tan\frac{x}{2}, direct differentiation gives tan⁡π4+π4sec⁡2π4=1+π2\tan\frac{\pi}{4}+\frac{\pi}{4}\sec^2\frac{\pi}{4}=1+\frac{\pi}{2}. ✓

06

On the exam

How this topic is marked

  • Check every sign before moving on: the minus belongs to cos⁡\cos, cot⁡\cot and csc⁡\csc, and a quotient rule written with uv′−u′vuv'-u'v on top loses the whole question.

  • A limit like lim⁡h→0sin⁡(a+h)−sin⁡ah\lim_{h\to0}\frac{\sin(a+h)-\sin a}{h} is cos⁡a\cos a in disguise — name ff and aa and answer in one line instead of expanding.

  • Simplify with sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 only after the rule is applied correctly, give exact values (32\frac{\sqrt3}{2}, not 0.8660.866), and keep the calculator in radians.

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