MATH 136
Cscx and Cotx
1. Derive the derivatives of
y = csc x =
1
= (sin x)−1
sin x
Then,
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dy
cos x
1
cos x
= −(sin x)−2 × cos x = − 2 = −
×
= −csc x cot x
dx
sin x sin x
sin x
Thus,
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d(csc x)
= −csc x cot x
dx
y = cot x =
cos x
sin x
Then,
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dy −sin x × sin x − cos x × cos x
sin 2 x + cos2 x
1
=
=
−
= − 2 = −csc 2 x
2
2
dx
sin x
sin x
sin x
Thus,
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d(cot x)
= −csc 2 x
dx
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2. Evaluate f ′(x) :
(a) f (x) = ln( csc x
)
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f ′(x) =
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csc x
1
×
× (−csc x cot x) = −cot x
csc x
csc x
(b) f (x) = cot 2 (4 π x)
(
)
f ′(x) = 2(cot(4 π x)) × −csc 2 (4 π x) × 4π
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= −8π cot(4 π x) csc 2 (4 π x)
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(c) f (x) = cot(3 x) × csc(3x)
f ′(x) = −csc 2 (3x) × 3 × csc(3x) + cot(3x) × (−csc(3x)cot(3x)) × 3
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= −3csc 3 (3x) − 3cot 2 (3x) csc(3x)
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(d) f (x) =
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f ′(x) =
cot 4 (2x)
csc 2 (2x)
(
)
4 cot 3 (2x) × −csc 2 (2x) × 2 × csc 2 (2x) − cot 4 (2x) × 2csc(2x)(−csc(2x)cot(2x)) × 2
= −8cot 3 (2x) +
csc 4 (2x)
4 cot 5 (2x)
csc 2 (2x)
(Can also simplify and re-write the original function as cos2 (2x) × cot 2 (2x) )
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