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 → 0 sin θ θ = 1 \lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}=1 lim θ → 0 θ sin θ = 1
The quotient is even, so treat θ → 0 + \theta\to 0^{+} θ → 0 + . On the unit circle, sector C O D COD C O D ≤ △ C O B \triangle COB △ C O B ≤ sector A O B AOB A O B yields
cos θ ≤ sin θ θ ≤ 1 cos θ → θ → 0 + 1. \cos\theta\le\frac{\sin\theta}{\theta}\le\frac{1}{\cos\theta}\xrightarrow{\theta\to 0^{+}}1. cos θ ≤ θ sin θ ≤ cos θ 1 θ → 0 + 1.
Fig. 17-11 — sector COD ≤ △COB ≤ sector AOB
2. Prove D x ( sin x ) = cos x D_x(\sin x)=\cos x D x ( sin x ) = cos x
Δ y Δ x = cos x ⋅ sin Δ x Δ x + sin x ⋅ cos Δ x − 1 Δ x → 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. Δ x Δ y = cos x ⋅ Δ x sin Δ x + sin x ⋅ Δ x cos Δ x − 1 → cos x .
3. Prove tan ′ = sec 2 \tan'={\sec}^{2} tan ′ = sec 2 and sec ′ = tan sec \sec'=\tan\sec sec ′ = tan sec
(a) Quotient rule on sin / cos \sin/\cos sin / cos .
(b) Two routes: reciprocal rule on 1 / cos 1/\cos 1/ cos , or differentiate tan 2 x + 1 = sec 2 x \tan^{2}x+1=\sec^{2}x tan 2 x + 1 = sec 2 x :
2 tan x sec 2 x = 2 sec x ⋅ D x ( sec x ) ⟹ D x ( sec x ) = tan x sec x . 2\tan x\sec^{2}x=2\sec x\cdot D_x(\sec x)\implies D_x(\sec x)=\tan x\sec x. 2 tan x sec 2 x = 2 sec x ⋅ D x ( sec x ) ⟹ D x ( sec x ) = tan x sec x .
4–5. Period of tan \tan tan and tan ( u − v ) \tan(u-v) tan ( u − v )
sin ( x + π ) = − sin x \sin(x+\pi)=-\sin x sin ( x + π ) = − sin x , cos ( x + π ) = − cos x \cos(x+\pi)=-\cos x cos ( x + π ) = − cos x ⇒ tan ( x + π ) = tan x \tan(x+\pi)=\tan x tan ( x + π ) = tan x .
Divide sin ( u − v ) / cos ( u − v ) \sin(u-v)/\cos(u-v) sin ( u − v ) / cos ( u − v ) by cos u cos v \cos u\cos v cos u cos v :
tan ( u − v ) = tan u − tan v 1 + tan u tan v . \tan(u-v)=\frac{\tan u-\tan v}{1+\tan u\tan v}. tan ( u − v ) = 1 + tan u tan v tan u − tan v .
(The plus formula flips signs: ( tan u + tan v ) / ( 1 − tan u tan v ) (\tan u+\tan v)/(1-\tan u\tan v) ( tan u + tan v ) / ( 1 − tan u tan v ) .)
6. Nested chain-rule derivatives
Book. 2 cos 7 x 2\cos 7x 2 cos 7 x , sin 3 ( 2 x ) \sin^{3}(2x) sin 3 ( 2 x ) , tan ( 5 x ) \tan(5x) tan ( 5 x ) , sec ( 1 / x ) \sec(1/x) sec ( 1/ x ) .
Nested trig derivatives Outer power / trig × inner chain 换一批(3 题)
(1) Differentiate: (a)
2 cos 7 x 2\cos 7x 2 cos 7 x (b)
sin 3 ( 2 x ) \sin^{3}(2x) sin 3 ( 2 x ) (c)
tan ( 5 x ) \tan(5x) tan ( 5 x ) (d)
sec ( 1 / x ) \sec(1/x) sec ( 1/ x ) .
显示解答
(a)
− 14 sin 7 x -14\sin 7x − 14 sin 7 x .
(b)
6 sin 2 ( 2 x ) cos ( 2 x ) 6\sin^{2}(2x)\cos(2x) 6 sin 2 ( 2 x ) cos ( 2 x ) .
(c)
5 sec 2 ( 5 x ) 5\sec^{2}(5x) 5 sec 2 ( 5 x ) .
(d)
− ( 1 / x 2 ) tan ( 1 / x ) sec ( 1 / x ) -(1/x^{2})\tan(1/x)\sec(1/x) − ( 1/ x 2 ) tan ( 1/ x ) sec ( 1/ x ) .
See steps.
(2) Differentiate
y = cot ( 1 − 2 x 2 ) y=\cot(1-2x^{2}) y = cot ( 1 − 2 x 2 ) .
显示解答
y ′ = − csc 2 ( 1 − 2 x 2 ) ⋅ ( − 4 x ) = 4 x csc 2 ( 1 − 2 x 2 ) y'=-\csc^{2}(1-2x^{2})\cdot(-4x)=4x\csc^{2}(1-2x^{2}) y ′ = − csc 2 ( 1 − 2 x 2 ) ⋅ ( − 4 x ) = 4 x csc 2 ( 1 − 2 x 2 ) .
y ′ = 4 x csc 2 ( 1 − 2 x 2 ) y'=4x\csc^{2}(1-2x^{2}) y ′ = 4 x csc 2 ( 1 − 2 x 2 ) .
(3) Differentiate
y = cot ( 1 − 2 x 2 ) y=\cot(1-2x^{2}) y = cot ( 1 − 2 x 2 ) .
显示解答
y ′ = − csc 2 ( 1 − 2 x 2 ) ⋅ ( − 4 x ) = 4 x csc 2 ( 1 − 2 x 2 ) y'=-\csc^{2}(1-2x^{2})\cdot(-4x)=4x\csc^{2}(1-2x^{2}) y ′ = − csc 2 ( 1 − 2 x 2 ) ⋅ ( − 4 x ) = 4 x csc 2 ( 1 − 2 x 2 ) .
y ′ = 4 x csc 2 ( 1 − 2 x 2 ) y'=4x\csc^{2}(1-2x^{2}) y ′ = 4 x csc 2 ( 1 − 2 x 2 ) .
7. Solve cos x = 1 2 \cos x=\tfrac12 cos x = 2 1 (and kin)
Unit-circle points with x x x -coordinate 1 / 2 1/2 1/2 : angles π / 3 \pi/3 π /3 and 5 π / 3 5\pi/3 5 π /3 in [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] ; then add 2 π n 2\pi n 2 π n .
Trig equations — general solution Principal solutions + period 换一批(2 题)
(1) Find all solutions of
cos x = 1 / 2 \cos x=1/2 cos x = 1/2 .
显示解答
On
[ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] :
x = π / 3 x=\pi/3 x = π /3 and
x = 5 π / 3 x=5\pi/3 x = 5 π /3 .
General:
π / 3 + 2 π n \pi/3+2\pi n π /3 + 2 π n and
5 π / 3 + 2 π n 5\pi/3+2\pi n 5 π /3 + 2 π n .
x = ± π / 3 + 2 π n x=\pm\pi/3+2\pi n x = ± π /3 + 2 π n (equiv. form OK).
(2) Find all solutions of
cos x = 0 \cos x=0 cos x = 0 .
显示解答
Odd multiples of
π / 2 \pi/2 π /2 .
x = ( 2 n + 1 ) π / 2 x=(2n+1)\pi/2 x = ( 2 n + 1 ) π /2 .
8. Limits via ( sin u ) / u → 1 (\sin u)/u\to 1 ( sin u ) / u → 1
Book. ( sin 5 x ) / ( 2 x ) → 5 / 2 (\sin 5x)/(2x)\to 5/2 ( sin 5 x ) / ( 2 x ) → 5/2 ; ( sin 3 x ) / ( sin 7 x ) → 3 / 7 (\sin 3x)/(\sin 7x)\to 3/7 ( sin 3 x ) / ( sin 7 x ) → 3/7 ; ( tan x ) / x → 1 (\tan x)/x\to 1 ( tan x ) / x → 1 .
Trig limits as x→0 Rewrite to sin u / u (radians) 换一批(3 题)
(1) Evaluate: (a)
lim x → 0 sin 5 x 2 x \lim_{x\to 0}\dfrac{\sin 5x}{2x} lim x → 0 2 x sin 5 x (b)
lim x → 0 sin 3 x sin 7 x \lim_{x\to 0}\dfrac{\sin 3x}{\sin 7x} lim x → 0 sin 7 x sin 3 x (c)
lim x → 0 tan x x \lim_{x\to 0}\dfrac{\tan x}{x} lim x → 0 x tan x .
显示解答
(a)
5 2 lim ( sin 5 x ) / ( 5 x ) = 5 / 2 \tfrac{5}{2}\lim(\sin 5x)/(5x)=5/2 2 5 lim ( sin 5 x ) / ( 5 x ) = 5/2 .
(b)
( sin 3 x ) / ( 3 x ) ⋅ ( 7 x ) / ( sin 7 x ) ⋅ 3 / 7 → 3 / 7 (\sin 3x)/(3x)\cdot(7x)/(\sin 7x)\cdot 3/7\to 3/7 ( sin 3 x ) / ( 3 x ) ⋅ ( 7 x ) / ( sin 7 x ) ⋅ 3/7 → 3/7 .
(c)
( sin x ) / x ⋅ 1 / cos x → 1 (\sin x)/x\cdot 1/\cos x\to 1 ( sin x ) / x ⋅ 1/ cos x → 1 .
(a)
5 / 2 5/2 5/2 ; (b)
3 / 7 3/7 3/7 ; (c)
1 1 1 .
(2) Evaluate
lim x → 0 sin ( 4 x ) 2 x \lim_{x\to 0}\dfrac{\sin(4x)}{2x} lim x → 0 2 x sin ( 4 x ) .
显示解答
a 2 lim sin ( 4 x ) 4 x = a 2 \dfrac{a}{2}\lim\dfrac{\sin(4x)}{4x}=\dfrac{a}{2} 2 a lim 4 x sin ( 4 x ) = 2 a .
(3) Evaluate
lim x → 0 sin ( 5 x ) 2 x \lim_{x\to 0}\dfrac{\sin(5x)}{2x} lim x → 0 2 x sin ( 5 x ) .
显示解答
a 2 lim sin ( 5 x ) 5 x = a 2 \dfrac{a}{2}\lim\dfrac{\sin(5x)}{5x}=\dfrac{a}{2} 2 a lim 5 x sin ( 5 x ) = 2 a .
9–12. Higher / implicit derivatives
Book 9. y = x sin x y=x\sin x y = x sin x ⇒ y ′ ′ ′ = − x cos x − 3 sin x y'''=-x\cos x-3\sin x y ′′′ = − x cos x − 3 sin x .
Book 10. y = tan 2 ( 3 x − 2 ) y=\tan^{2}(3x-2) y = tan 2 ( 3 x − 2 ) ⇒ first y ′ = 6 tan ( ⋯ ) sec 2 ( ⋯ ) y'=6\tan(\cdots)\sec^{2}(\cdots) y ′ = 6 tan ( ⋯ ) sec 2 ( ⋯ ) ; then product for y ′ ′ y'' y ′′ .
Book 11. y = sin ( x + y ) y=\sin(x+y) y = sin ( x + y ) ⇒ y