Math 220
October 9
I. Find dy/dx by implicit differentiation
1. y 3 = x2
2. y = sin−1 (x) + cos−1 (x) + tan−1 (x) + csc−1 (x) + sec−1 (x) + cot−1 (x)
√
√
3. 3 x + 2 y = 4
4. xy = cos(x)
5. x2 y 2 + y = x3 tan−1 (3x + 1)
6. tan(x + y) = y 3
7. cos(x + y 2 ) = y cos(y 3 x)
8. (x + y)3 + xey = sin(y)
9. cos(y/x) sin(xy 2 ) = 2y ex
2
II. Find the equation of the tangent line to the curve y 2 (y 2 − 4) = x2 (x2 − 5)
at (0,2), (1,1) and (1,0)
III Differentiate the function
1) y = ln(x)
2) y = x ln(x)
3) y = ln(ln(ln(x + 3)))
4) y = log2 (x)
5) y = log5 (log3 (x))
3
6) y = √
(x + 3)4 (x
+ 5)5√
(x7 + 6)9 (x3 + 4)2
√
3
7) y = 3x + 1 x2 + 5 6 x3 + 3x2 + 5(x4 + 6x + 1)5 8) y = x3x
1
9) y = cos(2x)x
10) y = xcsc(3x)
√
11) y = ln(x) x
12) y = xx
x
IV. Rates of change in natural science
1. A ball is thrown upward from the surface of Jupiter with velocity 130
m/s, its height after t seconds is h = 130t − 26t2 .
(a) When is velocity the ball half of its initial velocity?
(b) What is the ball maximum height?
(c) What is the ball’s acceleration?
(d) With what velocity does it hit the ground?
2. A particle moves with position function
s = t5 + t4 + 2t2 + 22t
(a) At what time is does the particle have velocity 35 m/s?
(b) What is the particle’s acceleration function
(c) When does the particle have jerk zero?
2
1
Solutions
I. Find dy/dx by implicit differentiation
1. y 3 = x2
Answer:
3y 2 (y 0 ) = 2x
2x
y0 = 2
3y
2x
dy
= 2
dx
3y
2. y = sin−1 (x) + cos−1 (x) + tan−1 (x) + csc−1 (x) + sec−1 (x) + cot−1 (x)
Answer:
dy
1
1
1
−1
1
−1
= y0 = √
−√
+
+ √
+ √
+
2
2
2
2
2
dx
1+x
1−x
1−x
x x − 1 x x − 1 1 + x2
√
√
3. 3 x + 2 y = 4
Answer:
d √
d
√
(3 x + 2 y) =
(4)
dx
dx
3
2
√ + √ y0 = 0
y
2 x
2
3
√ y0 = − √
y
2 x
√
dy
3 y
0
=y =− √
dx
4 x
3
4. xy = cos(x)
Answer:
d
d
(xy) =
(cos(x))
dx
dx
y + xy 0 = − sin(x)
y 0 x = − sin(x) − y
dy
− sin(x) − y
= y0 =
dx
x
5. x2 y 2 + y = x3 tan−1 (3x + 1)
Answer:
d 3
d 2 2
(x y + y) =
(x tan−1 (3x + 1))
dx
dx
3
1 + (3x + 1)2
3x3
− 2xy 2
2x2 yy 0 = 3x2 tan−1 (3x + 1) +
1 + (3x + 1)2
2xy 2 + 2x2 yy 0 = 3x2 tan−1 (3x + 1) + x3
3
3x
2
3x2 tan−1 (3x + 1) + 1+(3x+1)
2 − 2xy
dy
0
=y =
dx
2x2 y
6. tan(x + y) = y 3
Answer:
4
d
d 3
(tan(x + y)) =
(y )
dx
dx
sec2 (x + y)(1 + y 0 ) = 3y 2 y 0
sec2 (x + y) + y 0 sec2 (x + y) = 3y 2 y 0
y 0 sec2 (x + y) − 3y 2 y 0 = sec2 (x + y)
y 0 (sec2 (x + y) − 3y 2 ) = sec2 (x + y)
dy
sec2 (x + y)
= y0 =
dx
sec2 (x + y) − 3y 2
7. cos(x + y 2 ) = y cos(y 3 x)
Answer:
d
d
(cos(x + y 2 )) =
(y cos(y 3 x))
dx
dx
− sin(x + y 2 )(1 + 2yy 0 ) =
y 0 cos(y 3 x) + y(− sin(y 3 x)(3y 2 y 0 x + y 3 ))
− sin(x + y 2 ) − 2yy 0 sin(x + y 2 ) =
y 0 cos(y 3 x) − sin(y 3 x)3y 3 y 0 x − y 3 sin(y 3 x)
−2yy 0 sin(x + y 2 ) − y 0 cos(y 3 x) + y 0 sin(y 3 x)3y 3 x =
−y 3 sin(y 3 x) + sin(x + y 2 )
y 0 (−2y sin(x + y 2 ) − cos(y 3 x) + sin(y 3 x)3y 3 x) = −y 3 sin(y 3 x) + sin(x + y 2 )
−y 3 sin(y 3 x) + sin(x + y 2 )
dy
= y0 =
dx
−2y sin(x + y 2 ) − cos(y 3 x) + sin(y 3 x)3y 3 x
8. (x + y)3 + xey = sin(y)
Answer:
5
d
d
((x + y)3 + xey ) =
(sin(y))
dx
dx
3(x + y)2 (1 + y 0 ) + ey + xey y 0 = cos(y)y 0
3(x + y)2 + y 0 3(x + y)2 + ey + xey y 0 = cos(y)y 0
y 0 (3(x + y)2 + xey y 0 − cos(y)) = −ey − 3(x + y)2
9. cos(y/x) sin(xy 2 ) = 2y ex
Answer:
2
d
d
(cos(x + y 2 )) =
(y cos(y 3 x))
dx
dx
− sin(x + y 2 )(1 + 2yy 0 ) =
y 0 cos(y 3 x)+ − y sin(y 3 x)(3y 2 y 0 x + y 3 )
− sin(x + y 2 ) − 2yy 0 sin(x + y 2 ) =
y 0 cos(y 3 x) − y 0 sin(y 3 x)3y 3 x − y 4 sin(y 3 x)
−2yy 0 sin(x + y 2 ) − y 0 cos(y 3 x) + y 0 sin(y 3 x)3y 3 x = −y 4 sin(y 3 x) + sin(x + y 2 )
y 0 (−2y sin(x + y 2 ) − cos(y 3 x) + sin(y 3 x)3y 3 x) = −y 4 sin(y 3 x) + sin(x + y 2 )
−y 4 sin(y 3 x) + sin(x + y 2 )
dy
= y0 =
dx
−2y sin(x + y 2 ) − cos(y 3 x) + sin(y 3 x)3y 3 x
II. Find the equation of the tangent line to the curve y 2 (y 2 − 4) = x2 (x2 − 5)
at (0,2), (1,1) and (1,0)
Answer:
6
y 2 (y 2 − 4) = x2 (x2 − 5)
y 4 − 4y 2 = x4 − 5x2
d 4
d 4
(y − 4y 2 ) =
(x − 5x2 )
dx
dx
4y 3 y 0 − 8yy 0 = 4x3 − 10x
y 0 (4y 3 − 8y) = 4x3 − 10x
4x3 − 10x
d
= y0 =
dx
4y 3 − 8y
At (0, 2)
3 −10(0)
y 0 |(0,2) = 4(0)
=0
4(2)3 −8(2)
Tangent Line:
(y − 2) = 0(x − 0)
y=2
At (1, 1)
3 −10(1)
y 0 |(1,1) = 4(1)
= −6/(−4) = 3/2
4(1)3 −8(1)
Tangent Line:
(y − 1) = 3/2(x − 1)
− 12
y = 3x
2
At (1, 0)
3 −10(1)
y 0 |(1,0) = 4(1)
=
4(0)3 −8(0)
Tangent line:
x=1
−6
0
undefine
III Differentiate the function
1. y = ln(x)
Answer:
dy
1
=
dx
x
7
2. y = x ln(x)
1
dy
= ln(x) + x = ln(x) + 1
dx
x
3. y = ln(ln(ln(x + 3)))
dy
1
1
1
=
dx
ln(ln(x + 3) ln(x + 3) x + 3
4. y = log2 (x)
1 1
1
dy
=
=
dx
ln(2) x
x ln(2)
5. y = log5 (log3 (x))
dy
1
1
1 1
1
=
=
dx
ln(5) log3 (x) ln(3) x
ln(3) ln(5)x log3 (x)
6. y = (x + 3)4 (x3 + 5)5 (x7 + 6)9 (x3 + 4)2
y = (x + 3)4 (x3 + 5)5 (x7 + 6)9 (x3 + 4)2
ln(y) = ln((x + 3)4 (x3 + 5)5 (x7 + 6)9 (x3 + 4)2 )
ln(y) = ln((x + 3)4 ) + ln((x3 + 5)5 ) + ln((x7 + 6)9 ) + ln((x3 + 4)2 )
ln(y) = 4 ln(x + 3) + 5 ln(x3 + 5) + 9 ln(x7 + 6) + 2 ln((x3 + 4)
d
d
(ln(y)) =
(4 ln(x + 3) + 5 ln(x3 + 5) + 9 ln(x7 + 6) + 2 ln((x3 + 4))
dx
dx
y0
4
5(3x2 ) 9(7x6 )
2(3x2 )
=
+ 3
+ 7
+ 2
y
x + 3 x + 5 x + 6 3x + 4
4
15x2
63x6
6x2
0
y =
+
+
+
y
x + 3 x3 + 5 x7 + 6 3x2 + 4
0
y =
4
15x2
63x6
6x2
+ 3
+ 7
+ 2
x + 3 x + 5 x + 6 3x + 4
8
(x+3)4 (x3 +5)5 (x7 +6)9 (x3 +4)2
√
√
3x + 1 3 x2 + 5 6 x3 + 3x2 + 5(x4 + 6x + 1)5
√
√
√
3
6
y = 3x + 1 x2 + 5 x3 + 3x2 + 5(x4 + 6x + 1)5
√
√
√
3
6
ln(y) = ln( 3x + 1 x2 + 5 x3 + 3x2 + 5(x4 + 6x + 1)5 )
1
1
1
ln(y) = ln(3x + 1) + ln(x2 + 5) + ln(x3 + 3x2 + 5) + 5 ln(x4 + 6x + 1)
2
3
6
d
d 1
1
1
2
(ln(y)) =
( ln(3x + 1) + ln(x + 5) + ln(x3 + 3x2 + 5) + 5 ln(x4 + 6x + 1))
dx
dx 2
3
6
y0
1 3
1 2x
1 3x2 + 6x
5(4x3 + 6)
=
+
+
+
y
2 3x + 1 3 x2 + 5 6 x3 + 3x2 + 5 x4 + 6x + 1
2x
x2 + 2x
20x3 + 30
3
0
+
+
+
y
y =
6x + 2 3x2 + 15 2x3 + 6x2 + 10 x4 + 6x + 1
3
2x
x2 + 2x
20x3 + 30
0
y =
+
+
+
6x + 2 3x2 + 15 2x3 + 6x2 + 10 x4 + 6x + 1
√
√
√
3
6
· 3x + 1 x2 + 5 x3 + 3x2 + 5(x4 + 6x + 1)5
7. y =
√
8. y = x3x
Answer:
y = x3x
ln(y) = ln(x3x )
ln(y) = 3x ln(x)
d
d
(ln(y)) =
(3x ln(x))
dx
dx
y0
1
= 3 ln(x) + 3x
y
x
0
y = (3 ln(x) + 3)y
y 0 = (3 ln(x) + 3)x3x
9. y = cos(2x)x
Answer:
9
y = (cos(2x))x
ln(y) = ln(cos(2x)x )
ln(y) = x ln(cos(2x))
d
d
(ln(y)) =
x ln(cos(2x)))
dx
dx
− sin(2x)2
y0
= ln(cos(2x)) + x
y
cos(2x)
0
y = ln(cos(2x)) − 2x tan(2x)
10. y = xcsc(3x)
Answer:
y = xcsc(3x)
ln(y) = ln(xcsc(3x) )
ln(y) = csc(3x) ln(x)
d
d
(ln(y)) =
(csc(3x) ln(x))
dx
dx
y0
csc(3x)
= − csc(3x) cot(3x)3 ln(x) +
y
x
csc(3x)
0
y = −3 csc(3x) cot(3x) ln(x) +
y
x
csc(3x)
0
y = −3 csc(3x) cot(3x) ln(x) +
xcsc(3x)
x
√
11. y = ln(x)
Answer:
x
10
√
y = ln(x)
x
√
ln(y) = ln(ln(x) x )
√
ln(y) = x ln(ln(x))
d
d√
ln(y) =
x ln(ln(x))
dx
dx
√ 1 1
1
y 0 = ( √ ln(ln(x)) + x
)y
ln(x) x
2 x
√
√ 1 1
1
) ln(x) x
y 0 = ( √ ln(ln(x)) + x
ln(x) x
2 x
x
12. y = xx
Answer:
y = xx
x
x
ln(y) = ln(xx )
ln(y) = xx ln(x)
ln(ln(y)) = ln(xx ln(x))
d
d
ln(ln(y)) =
(x ln(x) + ln(x)))
dx
dx
1 y0
1
= ln(x) + 1 +
ln(y) y
x
1
y 0 = (ln(x) + 1 + )y ln(y)
x
1 x
y 0 = (ln(x) + 1 + )xx ln(x)xx
x
IV. Rates of change in natural science
1. A ball is thrown upward from the surface of Jupiter with velocity 130
m/s, its height after t seconds is h = 130t − 26t2 .
(a) When is velocity of the ball half of its initial velocity?
Answer:
11
v(t) = h0 (t) = 130 − 52t
Solve for t:
130
= 130 − 52t
2
=⇒ t = 130
= 54
52t = 130
2
104
At t = 1.25 the velocity of the ball half of its initial velocity.
(b) What is the ball’s maximum height? Answer:
The ball will reach its maximum height when v(t) = 0
= 2.5
v(t) = 130 − 52t = 0 =⇒ t = 130
52
h(2.5) = 130(2.5) − 26(2.5)2 = 162.5
Also note that we can rewrite h,
h(t) = −26t2 + 130t = −26(t2 − 5t + 6.25 − 6.25)
= −26(t − 2.5)2 + −26(−6.25)
= −26(t − 2.5)2 + 162.5
So we can see the vertex is at (2.5, 162.5)
So the maximum height is 162.5 feet.
(c) What is the ball’s acceleration?
a(t) = v 0 (t) = −52
The acceleration is −52 feet/second2 .
(d) With what velocity does it hit the ground?
It hits the ground when h(t) = 0
h(t) = 26t(5 − t)
h(t) = 0 =⇒ t = 0, 5
The ball will hit the ground after 5 seconds.
v(5) = 130 − 52(5) = −130
2. A particle moves with position function
s = t5 + t4 + 2t2 + 22t
(a) At what time is does the particle have velocity 35 m/s? Answer:
v(t) = 5t4 + 4t3 + 4t + 22
v(t) = 35 =⇒ t = 1
After 1 second.
(b) What is the particle’s acceleration function? a(t) = 20t3 + 13t + 4
12
(c) When does the particle have jerk zero? Jerk is the derivative of
acceleration.
j(t) = 60t2 + 13
13