SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS Calculus I – Worksheet #19 Review for Test 4 - Derivatives d ( x2 ) 2x y = x2 + 5 ⇒ y 2 = x2 + 5 1. = = 2 x3 d (ln x) 1 2. d ( x 2 + 5) 2 x 1 = 3= x 4 4x 2x d (x )
3. Simplify: e 4ln x = eln x = x 4 4
y = 3x +1 ⇒ ln y = ( x + 1) ln 3 ⇒ 4.
y = ( ln x ) ⇒ ln y = x • ln[ln x] x
5.
7.
1
=
y = cos x sin x ⇒ y ' = sin x(− sin x) + cos x(cos x) 8.
= − sin 2 x + cos 2 x = cos 2 x
4( x 2 + x)
=
4x + 2 2 x2 + x
=
4 x2 + 4 x 2x + 1 x2 + x
y = cos3 5 x = (cos 5 x)3 y ' = 3(cos 5 x) 2 (− sin 5 x)(5) = −15cos 2 5 x sin 5 x
D.
f '( x) = 2e 2 x ⇒ f '(2) = 2e 4 ⇒ y − e4 = 2e 4 ( x − 2)
4x + 2
4x + 2
10.
1 1 + 2x y'= +2 = x x
f ( x) = e 2 x @(2, e 4 ) 11.
−1 1 6. y ' = (4 x 2 + 4 x) 2 (8 x + 4) = 2
1⎤ ⎡ x ⎢⎣ln[ln x] + x ⎥⎦ (ln x)
y = ln ( xe 2 x ) = ln x + ln e 2 x = ln x + 2 x 9.
dy = (ln 3) y = (ln 3) • 3x +1 dx y = 4 x 2 + 4 x = (4 x 2 + 4 x) 2
1 dy 1 1 1 = ln[ln x](1) + x • • = ln[ln x] + ⇒ y dx x ln x x dy ⎡ 1⎤ = ⎢ ln[ln x] + ⎥ y = dx ⎣ x⎦
1 dy = ln 3 ⇒ y dx
d [ln( xe x )] d [ln x + x] 1 = = +1 dx dx x
12. y = ln(6 x 2 − 3) ⇒ y ' =
12 x 4x = 2 6 x − 3 2 x2 −1
14.
13. e 4 x + 2ln x = e 4 x • eln x = x 2 e 4 x 2
y = ln sin 3x ⇒ y ' =
1 • 3cos 3x = 3cot 3x sin 3x
. y = 2cos x ⇒ ln y = cos x(ln 2) ⇒ 15.
17.
1 dy = (ln 2)(− sin x) ⇒ y dx
dy = (− ln 2)(sin x) y ⇒ − ln 2(sin x) • 2cos x dx y = x 2 sec 4 x ⇒ (sec 4 x)(2 x) + x 2 sec 4 x tan 4 x(4) = 2 x sec 4 x(1 + 2 x tan 4 x)
16.
⎛ x2 y = ln ⎜ 6 x ⎝e y'=
⎞ 2 6x ⎟ = ln x − ln e = 2 ln x − 6 x ⎠
2 2 − 6x −6 = x x
d ( x 3 + 2 x 2 ) 3 x 2 + 4 x −(3 x 2 + 4 x) 18. = = d (cos x ) − sin x sin x
SOLUTIONS SOLUTIONS SOLUTIONS y = x cot 2 x ⇒ ln y = cot 2 x(ln x) ⇒
1 dy 1 = ln x(−2 csc 2 2 x) + cot 2 x • ⇒ y dx x cot 2 x ⎤ 19. dy ⎡ = ⎢ −2(ln x)(csc 2 x) + y dx ⎣ x ⎥⎦ cot 2 x ⎤ cot 2 x ⎡ x = ⎢ −2(ln x)(csc 2 x) + x ⎥⎦ ⎣
.
SOLUTIONS SOLUTIONS SOLUTIONS 20. y = cos 2 (3x) + sin 2 (3x) = 1 ⇒ y ' = 0
