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17b. Differentiation of Trig Functions — Practice

Drill sheet for 17. Differentiation of Trig Functions. Book-style solved ideas rewritten as guided proofs and regenerable worksheets — not a verbatim dump.


1. Prove lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}=1​

The quotient is even, so treat θ→0+\theta\to 0^{+}. On the unit circle, sector CODCOD ≤ △COB\triangle COB ≤ sector AOBAOB yields

cos⁡θ≤sin⁡θθ≤1cos⁡θ→θ→0+1.\cos\theta\le\frac{\sin\theta}{\theta}\le\frac{1}{\cos\theta}\xrightarrow{\theta\to 0^{+}}1.
Fig. 17-11 — sector COD ≤ △COB ≤ sector AOB

2. Prove Dx(sin⁡x)=cos⁡xD_x(\sin x)=\cos x​

ΔyΔx=cos⁡x⋅sin⁡ΔxΔx+sin⁡x⋅cos⁡Δx−1Δx→cos⁡x.\frac{\Delta y}{\Delta x} =\cos x\cdot\frac{\sin\Delta x}{\Delta x} +\sin x\cdot\frac{\cos\Delta x-1}{\Delta x} \to\cos x.

3. Prove tan⁡′=sec⁡2\tan'={\sec}^{2} and sec⁡′=tan⁡sec⁡\sec'=\tan\sec​

(a) Quotient rule on sin⁡/cos⁡\sin/\cos.

(b) Two routes: reciprocal rule on 1/cos⁡1/\cos, or differentiate tan⁡2x+1=sec⁡2x\tan^{2}x+1=\sec^{2}x:

2tan⁡xsec⁡2x=2sec⁡x⋅Dx(sec⁡x)  ⟹  Dx(sec⁡x)=tan⁡xsec⁡x.2\tan x\sec^{2}x=2\sec x\cdot D_x(\sec x)\implies D_x(\sec x)=\tan x\sec x.

4–5. Period of tan⁡\tan and tan⁡(u−v)\tan(u-v)​

sin⁡(x+π)=−sin⁡x\sin(x+\pi)=-\sin x, cos⁡(x+π)=−cos⁡x\cos(x+\pi)=-\cos x ⇒ tan⁡(x+π)=tan⁡x\tan(x+\pi)=\tan x.

Divide sin⁡(u−v)/cos⁡(u−v)\sin(u-v)/\cos(u-v) by cos⁡ucos⁡v\cos u\cos v:

tan⁡(u−v)=tan⁡u−tan⁡v1+tan⁡utan⁡v.\tan(u-v)=\frac{\tan u-\tan v}{1+\tan u\tan v}.

(The plus formula flips signs: (tan⁡u+tan⁡v)/(1−tan⁡utan⁡v)(\tan u+\tan v)/(1-\tan u\tan v).)


6. Nested chain-rule derivatives​

Book. 2cos⁡7x2\cos 7x, sin⁡3(2x)\sin^{3}(2x), tan⁡(5x)\tan(5x), sec⁡(1/x)\sec(1/x).

Nested trig derivatives

Outer power / trig × inner chain
(1)Differentiate: (a) 2cos⁡7x2\cos 7x (b) sin⁡3(2x)\sin^{3}(2x) (c) tan⁡(5x)\tan(5x) (d) sec⁡(1/x)\sec(1/x).
(a) −14sin⁡7x-14\sin 7x.
(b) 6sin⁡2(2x)cos⁡(2x)6\sin^{2}(2x)\cos(2x).
(c) 5sec⁡2(5x)5\sec^{2}(5x).
(d) −(1/x2)tan⁡(1/x)sec⁡(1/x)-(1/x^{2})\tan(1/x)\sec(1/x).
See steps.
(2)Differentiate y=cot⁡(1−2x2)y=\cot(1-2x^{2}).
y′=−csc⁡2(1−2x2)⋅(−4x)=4xcsc⁡2(1−2x2)y'=-\csc^{2}(1-2x^{2})\cdot(-4x)=4x\csc^{2}(1-2x^{2}).
y′=4xcsc⁡2(1−2x2)y'=4x\csc^{2}(1-2x^{2}).
(3)Differentiate y=cot⁡(1−2x2)y=\cot(1-2x^{2}).
y′=−csc⁡2(1−2x2)⋅(−4x)=4xcsc⁡2(1−2x2)y'=-\csc^{2}(1-2x^{2})\cdot(-4x)=4x\csc^{2}(1-2x^{2}).
y′=4xcsc⁡2(1−2x2)y'=4x\csc^{2}(1-2x^{2}).

7. Solve cos⁡x=12\cos x=\tfrac12 (and kin)​

Unit-circle points with xx-coordinate 1/21/2: angles π/3\pi/3 and 5π/35\pi/3 in [0,2π][0,2\pi]; then add 2πn2\pi n.

Trig equations — general solution

Principal solutions + period
(1)Find all solutions of cos⁡x=1/2\cos x=1/2.
On [0,2π][0,2\pi]: x=π/3x=\pi/3 and x=5π/3x=5\pi/3.
General: π/3+2πn\pi/3+2\pi n and 5π/3+2πn5\pi/3+2\pi n.
x=±π/3+2πnx=\pm\pi/3+2\pi n (equiv. form OK).
(2)Find all solutions of cos⁡x=0\cos x=0.
Odd multiples of π/2\pi/2.
x=(2n+1)π/2x=(2n+1)\pi/2.

8. Limits via (sin⁡u)/u→1(\sin u)/u\to 1​

Book. (sin⁡5x)/(2x)→5/2(\sin 5x)/(2x)\to 5/2; (sin⁡3x)/(sin⁡7x)→3/7(\sin 3x)/(\sin 7x)\to 3/7; (tan⁡x)/x→1(\tan x)/x\to 1.

Trig limits as x→0

Rewrite to sin u / u (radians)
(1)Evaluate: (a) lim⁡x→0sin⁡5x2x\lim_{x\to 0}\dfrac{\sin 5x}{2x} (b) lim⁡x→0sin⁡3xsin⁡7x\lim_{x\to 0}\dfrac{\sin 3x}{\sin 7x} (c) lim⁡x→0tan⁡xx\lim_{x\to 0}\dfrac{\tan x}{x}.
(a) 52lim⁡(sin⁡5x)/(5x)=5/2\tfrac{5}{2}\lim(\sin 5x)/(5x)=5/2.
(b) (sin⁡3x)/(3x)⋅(7x)/(sin⁡7x)⋅3/7→3/7(\sin 3x)/(3x)\cdot(7x)/(\sin 7x)\cdot 3/7\to 3/7.
(c) (sin⁡x)/x⋅1/cos⁡x→1(\sin x)/x\cdot 1/\cos x\to 1.
(a) 5/25/2; (b) 3/73/7; (c) 11.
(2)Evaluate lim⁡x→0sin⁡(4x)2x\lim_{x\to 0}\dfrac{\sin(4x)}{2x}.
a2lim⁡sin⁡(4x)4x=a2\dfrac{a}{2}\lim\dfrac{\sin(4x)}{4x}=\dfrac{a}{2}.
42\dfrac{4}{2}.
(3)Evaluate lim⁡x→0sin⁡(5x)2x\lim_{x\to 0}\dfrac{\sin(5x)}{2x}.
a2lim⁡sin⁡(5x)5x=a2\dfrac{a}{2}\lim\dfrac{\sin(5x)}{5x}=\dfrac{a}{2}.
52\dfrac{5}{2}.

9–12. Higher / implicit derivatives​

Book 9. y=xsin⁡xy=x\sin x ⇒ y′′′=−xcos⁡x−3sin⁡xy'''=-x\cos x-3\sin x.

Book 10. y=tan⁡2(3x−2)y=\tan^{2}(3x-2) ⇒ first y′=6tan⁡(⋯ )sec⁡2(⋯ )y'=6\tan(\cdots)\sec^{2}(\cdots); then product for y′′y''.

Book 11. y=sin⁡(x+y)y=\sin(x+y) ⇒ y′=cos⁡(x+y)1−cos⁡(x+y)y'=\dfrac{\cos(x+y)}{1-\cos(x+y)}.

Book 12. sin⁡y+cos⁡x=1\sin y+\cos x=1 ⇒ y′=sin⁡x/cos⁡yy'=\sin x/\cos y; differentiate again for y′′y''.

Higher and implicit

Repeated product / solve for y′
(1)If y=xsin⁡xy=x\sin x, find y′′′y'''.
y′=xcos⁡x+sin⁡xy'=x\cos x+\sin x.
y′′=−xsin⁡x+2cos⁡xy''=-x\sin x+2\cos x.
y′′′=−xcos⁡x−3sin⁡xy'''=-x\cos x-3\sin x.
y′′′=−xcos⁡x−3sin⁡xy'''=-x\cos x-3\sin x.
(2)If y=tan⁡2(3x−2)y=\tan^{2}(3x-2), find y′y' (first derivative only).
y′=2tan⁡(3x−2)sec⁡2(3x−2)⋅3=6tan⁡(3x−2)sec⁡2(3x−2)y'=2\tan(3x-2)\sec^{2}(3x-2)\cdot 3=6\tan(3x-2)\sec^{2}(3x-2).
y′=6tan⁡(3x−2)sec⁡2(3x−2)y'=6\tan(3x-2)\sec^{2}(3x-2).
(3)If sin⁡y+cos⁡x=1\sin y+\cos x=1, find y′y' (first derivative).
cos⁡y y′−sin⁡x=0\cos y\,y'-\sin x=0 ⇒ y′=sin⁡x/cos⁡yy'=\sin x/\cos y.
y′=sin⁡xcos⁡yy'=\dfrac{\sin x}{\cos y}.

Altitude 22 mi, ground distance x=2cot⁡θx=2\cot\theta, dx/dt=−240dx/dt=-240 mi/h. At θ=30°\theta=30°,

dxdt=−2csc⁡2θdθdt  ⟹  dθdt=30 rad/h.\frac{dx}{dt}=-2\csc^{2}\theta\frac{d\theta}{dt}\implies\frac{d\theta}{dt}=30\text{ rad/h}.
Fig. 17-12 — x = 2 cot θ

Related rates with cot / csc

Differentiate the geometric relation in t
(1)A 5050 ft pole; sun elevation θ=45∘\theta=45^{\circ} with dθ/dt=−1/4d\theta/dt=-1/4 rad/h. Shadow length s=50cot⁡θs=50\cot\theta. How fast is the shadow lengthening?
ds/dt=−50csc⁡2θ dθ/dtds/dt=-50\csc^{2}\theta\,d\theta/dt.
At 45∘45^{\circ}, csc⁡=2\csc=\sqrt{2}, so −50⋅2⋅(−1/4)=25-50\cdot 2\cdot(-1/4)=25 ft/h.
2525 ft/h.
(2)A 5050 ft pole; sun elevation θ=45∘\theta=45^{\circ} with dθ/dt=−1/4d\theta/dt=-1/4 rad/h. Shadow length s=50cot⁡θs=50\cot\theta. How fast is the shadow lengthening?
ds/dt=−50csc⁡2θ dθ/dtds/dt=-50\csc^{2}\theta\,d\theta/dt.
At 45∘45^{\circ}, csc⁡=2\csc=\sqrt{2}, so −50⋅2⋅(−1/4)=25-50\cdot 2\cdot(-1/4)=25 ft/h.
2525 ft/h.

14. Sketch f(x)=sin⁡x+cos⁡xf(x)=\sin x+\cos x​

Period 2π2\pi. f′=cos⁡−sin⁡=0f'=\cos-\sin=0 ⇒ tan⁡x=1\tan x=1 ⇒ x=π/4x=\pi/4 (max 2\sqrt{2}), x=5π/4x=5\pi/4 (min −2-\sqrt{2}). Inflection where tan⁡x=−1\tan x=-1: 3π/43\pi/4, 7π/47\pi/4.

Fig. 17-13 — y = sin x + cos x

15. Sketch f(x)=cos⁡x−cos⁡2xf(x)=\cos x-\cos^{2}x​

Even and 2π2\pi-periodic; work on [0,π][0,\pi]. f′=sin⁡x(2cos⁡x−1)=0f'=\sin x(2\cos x-1)=0 ⇒ x=0,π/3,πx=0,\pi/3,\pi. Classify: mins at 00 and π\pi; max at (π/3,1/4)(\pi/3,1/4). Inflections from f′′=0f''=0 (quadratic in cos⁡x\cos x).

Fig. 17-14 — y = cos x − cos² x

Trig extrema / absolute extrema

Critical numbers + endpoint table
(1)Critical numbers of f(x)=sin⁡x+cos⁡xf(x)=\sin x+\cos x on [0,2π][0,2\pi] and classification.
tan⁡x=1\tan x=1 ⇒ π/4\pi/4 (max 2\sqrt{2}), 5π/45\pi/4 (min −2-\sqrt{2}).
Max (π/4,2)(\pi/4,\sqrt{2}); min (5π/4,−2)(5\pi/4,-\sqrt{2}).
(2)Critical numbers of f(x)=sin⁡x+cos⁡xf(x)=\sin x+\cos x on [0,2π][0,2\pi] and classification.
tan⁡x=1\tan x=1 ⇒ π/4\pi/4 (max 2\sqrt{2}), 5π/45\pi/4 (min −2-\sqrt{2}).
Max (π/4,2)(\pi/4,\sqrt{2}); min (5π/4,−2)(5\pi/4,-\sqrt{2}).

16. Absolute extrema of sin⁡x+x\sin x+x on [0,2π][0,2\pi]​

f′=cos⁡x+1=0f'=\cos x+1=0 ⇒ x=πx=\pi. Table: f(0)=0f(0)=0, f(π)=πf(\pi)=\pi, f(2π)=2πf(2\pi)=2\pi ⇒ abs min 00 at 00; abs max 2π2\pi at 2π2\pi.


17–18. Angle between lines / curves​

Book 17. m1=1m_1=1, m2=−3m_2=-3:

tan⁡(α2−α1)=m2−m11+m1m2=2(≈63.4∘).\tan(\alpha_2-\alpha_1)=\frac{m_2-m_1}{1+m_1 m_2}=2\quad(\approx 63.4^\circ).

Book 18. At (1,1)(1,1) on y=x2y=x^{2} and y=xy=\sqrt{x}: slopes 22 and 1/21/2; tan⁡α=3/4\tan\alpha=3/4 (≈36.9∘\approx 36.9^\circ).

Fig. 17-15 — intersection angle from slopes

tan(α₂−α₁) from slopes

Use the tan(u−v) formula
(1)Angle between y=x+1y=x+1 (m1=1m_1=1) and y=−3x+5y=-3x+5 (m2=−3m_2=-3): find tan⁡(α2−α1)\tan(\alpha_2-\alpha_1).
tan⁡(α2−α1)=m2−m11+m1m2=−4−2=2\tan(\alpha_2-\alpha_1)=\dfrac{m_2-m_1}{1+m_1 m_2}=\dfrac{-4}{-2}=2.
tan⁡=2\tan=2 (≈63.4∘\approx 63.4^{\circ}).
(2)Slopes m1=1m_1=1, m2=0m_2=0: acute intersection angle?
tan⁡ϕ=1\tan\phi=1 ⇒ ϕ=π/4\phi=\pi/4.
π/4\pi/4.

Supplementary drills​

Period / amplitude / equations​

Book 19–22. Periodicity of cot⁡,sec⁡,csc⁡\cot,\sec,\csc; 5sin⁡(x/3)5\sin(x/3) has p=6πp=6\pi, f=1/3f=1/3, A=5A=5; cos⁡x=0\cos x=0 ⇒ (2n+1)π/2(2n+1)\pi/2; tan⁡x=1\tan x=1 ⇒ (4n+1)π/4(4n+1)\pi/4.

Amplitude / period / frequency

y = A sin(bx) readout
(1)For y=1.5sin⁡4xy=1.5\sin 4x: amplitude, period, frequency.
Amplitude ∣A∣=1.5|A|=1.5.
Period 2π/b=2π/4=π/22\pi/b=2\pi/4=\pi/2.
Frequency b=4b=4.
amp 1.51.5; p=π/2p=\pi/2; f=4f=4.
(2)For y=1.5cos(6x)y=1.5cos(6x): give amplitude, period, and frequency.
Amplitude ∣A∣=1.5|A|=1.5.
Period 2π/6=13π2\pi/6=\dfrac{1}{3}\pi.
Frequency b=6b=6.
amp 1.51.5; p=13πp=\dfrac{1}{3}\pi; f=6f=6.

Sketch sin⁡x2−cos⁡x\dfrac{\sin x}{2-\cos x}​

Domain all xx (2−cos⁡x≥12-\cos x\ge 1). Period 2π2\pi; use quotient rule for critical numbers.

Fig. 17-16 — y = sin x / (2 − cos x)

More differentiation (Book 25)​

Mixed differentiation

Product, quotient, and chain
(1)Differentiate y=sin⁡xcos⁡xy=\sin x\cos x.
Product: cos⁡xcos⁡x+sin⁡x(−sin⁡x)=cos⁡2x−sin⁡2x=cos⁡2x\cos x\cos x+\sin x(-\sin x)=\cos^{2}x-\sin^{2}x=\cos 2x.
y′=cos⁡2xy'=\cos 2x (or cos⁡2x−sin⁡2x\cos^{2}x-\sin^{2}x).
(2)Differentiate y=sin⁡xcos⁡xy=\sin x\cos x.
Product: cos⁡xcos⁡x+sin⁡x(−sin⁡x)=cos⁡2x−sin⁡2x=cos⁡2x\cos x\cos x+\sin x(-\sin x)=\cos^{2}x-\sin^{2}x=\cos 2x.
y′=cos⁡2xy'=\cos 2x (or cos⁡2x−sin⁡2x\cos^{2}x-\sin^{2}x).
(3)Differentiate y=sin⁡xcos⁡xy=\sin x\cos x.
Product: cos⁡xcos⁡x+sin⁡x(−sin⁡x)=cos⁡2x−sin⁡2x=cos⁡2x\cos x\cos x+\sin x(-\sin x)=\cos^{2}x-\sin^{2}x=\cos 2x.
y′=cos⁡2xy'=\cos 2x (or cos⁡2x−sin⁡2x\cos^{2}x-\sin^{2}x).

Basic chain compositions

sin(kx), tan(kx), …
(1)Differentiate: (a) y=sin⁡3xy=\sin 3x (b) y=cos⁡(x2)y=\cos(x^{2}) (c) y=tan⁡2xy=\tan 2x.
Dx(sin⁡3x)=cos⁡3x⋅3=3cos⁡3xD_x(\sin 3x)=\cos 3x\cdot 3=3\cos 3x.
Dx(cos⁡(x2))=−sin⁡(x2)⋅2x=−2xsin⁡(x2)D_x(\cos(x^{2}))=-\sin(x^{2})\cdot 2x=-2x\sin(x^{2}).
Dx(tan⁡2x)=sec⁡2(2x)⋅2=2sec⁡2(2x)D_x(\tan 2x)=\sec^{2}(2x)\cdot 2=2\sec^{2}(2x).
(a) 3cos⁡3x3\cos 3x; (b) −2xsin⁡(x2)-2x\sin(x^{2}); (c) 2sec⁡2(2x)2\sec^{2}(2x).
(2)Differentiate y=cos(5x)y=cos(5x).
Chain rule: outer trig derivative times 55.
y′=−5sin⁡5xy'=-5\sin 5x.

Limits (Book 26)​

lim⁡(sin⁡ax)/(sin⁡bx)=a/b\lim(\sin ax)/(\sin bx)=a/b; lim⁡sin⁡3(2x)/(xsin⁡2(3x))=8/9\lim\sin^{3}(2x)/(x\sin^{2}(3x))=8/9.

More trig limits

Factor constants into sin u / u
(1)Evaluate lim⁡x→0sin⁡(6x)sin⁡(7x)\lim_{x\to 0}\dfrac{\sin(6x)}{\sin(7x)}.
sin⁡(6x)6x⋅7xsin⁡(7x)⋅ab→ab\dfrac{\sin(6x)}{6x}\cdot\dfrac{7x}{\sin(7x)}\cdot\dfrac{a}{b}\to\dfrac{a}{b}.
67\dfrac{6}{7}.
(2)Evaluate lim⁡x→0tan⁡(4x)x\lim_{x\to 0}\dfrac{\tan(4x)}{x}.
sin⁡(4x)x⋅1cos⁡(4x)=a⋅sin⁡(4x)4x⋅1cos⁡(4x)→a\dfrac{\sin(4x)}{x}\cdot\dfrac{1}{\cos(4x)}=a\cdot\dfrac{\sin(4x)}{4x}\cdot\dfrac{1}{\cos(4x)}\to a.
44.

If x=Asin⁡kt+Bcos⁡ktx=A\sin kt+B\cos kt, then x′′=−k2xx''=-k^{2}x. For y=3sin⁡(2x+3)y=3\sin(2x+3), y′′+4y=0y''+4y=0. Sketch gallery on [0,2π][0,2\pi]: 12sin⁡2x\tfrac12\sin 2x, cos⁡2x−cos⁡x\cos^{2}x-\cos x, x−2sin⁡xx-2\sin x, etc. — find crits with y′=0y'=0 and classify.

Extrema / mono on trig graphs

Sign charts for y′ and y″
(1)Critical numbers of sec⁡x\sec x in (−π,π)(-\pi,\pi) where defined.
tan⁡xsec⁡x=0\tan x\sec x=0 ⇒ tan⁡x=0\tan x=0 ⇒ x=0x=0 (also ±π\pm\pi on a closed picture).
First-derivative test: relative min at 00.
Crit at x=0x=0 (relative min).
(2)Find inflection points of y=sin⁡xy=\sin x in [0,2π][0,2\pi].
y′′=−sin⁡x=0y''=-\sin x=0 ⇒ x=0,π,2πx=0,\pi,2\pi; sign of y′′y'' changes at each.
(0,0)(0,0), (π,0)(\pi,0), (2π,0)(2\pi,0).

Shadow of a 5050 ft pole with θ′=−1/4\theta'= -1/4 rad/h at 45∘45^\circ lengthens at 2525 ft/h. Implicit: tan⁡y=x2\tan y=x^{2}, cos⁡(xy)=2y\cos(xy)=2y.

Implicit y′

Differentiate both sides; solve for y′
(1)If y=sin⁡(x+y)y=\sin(x+y), solve for y′y'.
y′=cos⁡(x+y)(1+y′)y'=\cos(x+y)(1+y').
y′(1−cos⁡(x+y))=cos⁡(x+y)y'(1-\cos(x+y))=\cos(x+y).
y′=cos⁡(x+y)1−cos⁡(x+y)y'=\dfrac{\cos(x+y)}{1-\cos(x+y)}.
(2)If y=sin⁡(x+y)y=\sin(x+y), solve for y′y'.
y′=cos⁡(x+y)(1+y′)y'=\cos(x+y)(1+y').
y′(1−cos⁡(x+y))=cos⁡(x+y)y'(1-\cos(x+y))=\cos(x+y).
y′=cos⁡(x+y)1−cos⁡(x+y)y'=\dfrac{\cos(x+y)}{1-\cos(x+y)}.

Mixed review set​

Supplementary mixed set

Problems 6–31 style
(1)For y=3cos(5x)y=3cos(5x): give amplitude, period, and frequency.
Amplitude ∣A∣=3|A|=3.
Period 2π/5=25π2\pi/5=\dfrac{2}{5}\pi.
Frequency b=5b=5.
amp 33; p=25πp=\dfrac{2}{5}\pi; f=5f=5.
(2)Differentiate: (a) y=sin⁡3xy=\sin 3x (b) y=cos⁡(x2)y=\cos(x^{2}) (c) y=tan⁡2xy=\tan 2x.
Dx(sin⁡3x)=cos⁡3x⋅3=3cos⁡3xD_x(\sin 3x)=\cos 3x\cdot 3=3\cos 3x.
Dx(cos⁡(x2))=−sin⁡(x2)⋅2x=−2xsin⁡(x2)D_x(\cos(x^{2}))=-\sin(x^{2})\cdot 2x=-2x\sin(x^{2}).
Dx(tan⁡2x)=sec⁡2(2x)⋅2=2sec⁡2(2x)D_x(\tan 2x)=\sec^{2}(2x)\cdot 2=2\sec^{2}(2x).
(a) 3cos⁡3x3\cos 3x; (b) −2xsin⁡(x2)-2x\sin(x^{2}); (c) 2sec⁡2(2x)2\sec^{2}(2x).
(3)Differentiate y=sin⁡xy=\sqrt{\sin x} (where sin⁡x>0\sin x>0).
Chain: 12(sin⁡x)−1/2cos⁡x=cos⁡x2sin⁡x\dfrac{1}{2}(\sin x)^{-1/2}\cos x=\dfrac{\cos x}{2\sqrt{\sin x}}.
y′=cos⁡x2sin⁡xy'=\dfrac{\cos x}{2\sqrt{\sin x}}.
(4)Slopes m1=2m_1=2, m2=−1/2m_2=-1/2: are the lines perpendicular?
m1m2=2⋅(−1/2)=−1m_1 m_2=2\cdot(-1/2)=-1 ⇒ yes, 1+m1m2=01+m_1 m_2=0.
Yes — perpendicular.
(5)Evaluate lim⁡x→0tan⁡(4x)x\lim_{x\to 0}\dfrac{\tan(4x)}{x}.
sin⁡(4x)x⋅1cos⁡(4x)=a⋅sin⁡(4x)4x⋅1cos⁡(4x)→a\dfrac{\sin(4x)}{x}\cdot\dfrac{1}{\cos(4x)}=a\cdot\dfrac{\sin(4x)}{4x}\cdot\dfrac{1}{\cos(4x)}\to a.
44.
(6)Critical numbers of sec⁡x\sec x in (−π,π)(-\pi,\pi) where defined.
tan⁡xsec⁡x=0\tan x\sec x=0 ⇒ tan⁡x=0\tan x=0 ⇒ x=0x=0 (also ±π\pm\pi on a closed picture).
First-derivative test: relative min at 00.
Crit at x=0x=0 (relative min).

Self-check​

Quiz

Trig differentiation review

1 / 5

lim θ→0 (sin θ)/θ equals…

Pick an answer to see if you are right.

Further Reading​