Math 220 February 26 I. Find y/. 1. y = 1/x 2. y = sin(x) + cos(x) + tan

Math 220
February 26
I. Find y 0 .
1. y = 1/x3 + 5x4 + ex3 + x−π + eln(2)
2. y = sin(x) + cos(x) + tan(x) + csc(x) + sec(x) + cot(x) + ex + 3x
3. y = ln(x) + log2 (x) + log3 (3x + 3)
4. y 2 = x + 3
5. y = sin−1 (x)
6. y = cos−1 (x)
7. y = tan−1 (x)
8. y = sec−1 (x)
9. y = csc−1 (x)
10. y = cot−1 (x)
11. y = arctan(sin(x) + π)
12. cos(xy) = x + sin(y)
13. x2 + y 2 =
√
x + ln(3xy)
14. x = ey cos(x)
15. y = xx+1
16. (y + x)4 = xy 3 + x3 y
1
17.
y = ln
18. log3 (y) = sin(x) cos
19. tan(xy) =
x2 + 3
√
x + ex
x
y
x
x3 +y 4 +xy
√
20. y = (x + 2) x + 1 sin(x) cos(x) arctan(x)(x + 3)−2 e−5x
21. etan(ax+by) = sin(cx) + cos(dy)
22. y x = cos(2arcsin(x) )
√
23. 3y cos(y sin(x)) = tan−1 ( x + y + ex )
24. y = xy
II. Find an equation for the tangent line to the curve at the given point.
1. y 2 + x2 = 2xy, (1, 1)
2. y 3 = x + sin(x + π2 ), (0, 1)
3. y = ln(cos(x)), (0, 0)
x
4. y = x(x ) , (2, 16)
2
1
Solutions
I. Find y 0 .
1. y = 1/x3 + 5x4 + ex3 + x−π + eln(2)
Answer:
y 0 = −3x−4 + 20x3 + 3ex2 − πx−π−1
2. y = sin(x) + cos(x) + tan(x) + csc(x) + sec(x) + cot(x) + ex + 3x
Answer:
y 0 = cos(x)−sin(x)+sec2 (x)−csc(x) cot(x)+sec(x) tan(x)−csc2 (x)+ex +ln(3)3x
3. y = ln(x) + log2 (x) + log3 (3x + 3)
y0 =
1
3
1
+
+
x ln(2)x ln(3)(3x + 3)
4. y 2 = x + 3
Answer:
(y 2 )0 = (x + 3)0
2yy 0 = 1
1
y0 =
2y
5. y = sin−1 (x)
Answer:
3
(sin(y))0 = (x)0
cos(y)y 0 = 1
1
y0 =
cos(y)
1
y0 = p
1 − sin2 (x)
1
y0 = √
1 − x2
6. y = cos−1 (x)
Answer:
(cos(y))0 = (x)0
− sin(y)y 0 = 1
1
− sin(y)
−1
y0 = p
1 − cos2 (x)
−1
y0 = √
1 − x2
y0 =
7. y = tan−1 (x)
Answer:
4
(tan(y))0 = (x)0
sec2 (y)y 0 = 1
1
y0 =
sec2 (x)
1
y0 =
1 + tan2 (x)
1
y0 =
1 + x2
8. y = sec−1 (x)
Answer:
(sec(y))0 = (x)0
sec(y) tan(y)y 0 = 1
1
sec(y) tan(y)
1
y0 =
x tan(y)
1
y0 = p
2
x sec (y) − 1
1
y0 = √
x x2 − 1
y0 =
9. y = csc−1 (x)
Answer:
5
(csc(y))0 = (x)0
− csc(y) cot(y)y 0 = 1
−1
csc(y) cot(y)
−1
y0 =
x cot(y)
−1
y0 = p
x csc2 (y) − 1
−1
y0 = √
x x2 − 1
y0 =
10. y = cot−1 (x) Answer:
(cot(y))0 = (x)0
− csc2 (y)y 0 = 1
−1
y0 =
csc2 (y)
−1
y0 =
1 + cot2 (x)
−1
y0 =
1 + x2
11. y = arctan(sin(x) + π)
Answer:
1
(sin(x) + π)0
1 + (sin(x) + π)2
cos(x)
=
1 + (sin(x) + π)2
y0 =
6
12. cos(xy) = x + sin(y)
Answer:
(cos(xy))0 = (x + sin(y))0
− sin(xy)(y + xy 0 ) = 1 + cos(y)y 0
− sin(xy)y − sin(xy)xy 0 = 1 + cos(y)y 0
− sin(xy)xy 0 − cos(y)y 0 = 1 + sin(xy)y
y 0 (− sin(xy)x − cos(y)) = 1 + sin(xy)y
1 + sin(xy)y
y0 =
− sin(xy)x − cos(y)
13. x2 + y 2 =
Answer:
√
x + ln(3xy)
√
(x2 + y 2 )0 = ( x + ln(3xy))0
1
3y + 3xy 0
2x + 2yy 0 = √ +
3xy
2 x
1
1 y0
2x + 2yy 0 = √ + +
y
2 x x
0
y
1
1
2yy 0 − = √ + − 2x
y
2 x x
1
1
1
y 0 (2y + ) = √ + − 2x
y
2 x x
1
√
+ x1 − 2x
2 x
y0 =
2y + y1
14. x = ey cos(x)
Answer:
7
(x)0 = (ey cos(x) )0
1 = ey cos(x) (y 0 cos(x) − y sin(y))
1 = y 0 cos(x)ey cos(x) − y sin(y)ey cos(x)
y 0 cos(x)ey cos(x) = 1 + y sin(y)ey cos(x)
1 + y sin(y)ey cos(x)
cos(x)ey cos(x)
1 + xy sin(y)
y0 =
x cos(x)
y0 =
15. y = xx+1
Answer:
ln(y) = ln(xx+1 )
ln(y) = (x + 1) ln(x)
(ln(y))0 = ((x + 1) ln(x))0
x+1
y0
= ln(x) +
y
x
1
y 0 = (ln(x) + 1 + )y
x 1
xx+1
y 0 = ln(x) + 1 +
x
16. (y + x)4 = xy 3 + x3 y
Answer:
8
((y + x)4 )0 = (xy 3 + x3 y)0
4(y + x)3 (y 0 + x) = y 3 + x3y 2 y 0 + 3x2 y + x3 y 0
4(y + x)3 y 0 + 4x(y + x)3 = y 3 + 3xy 3 y 0 + 3x2 y + x3 y 0
4(y + x)3 y 0 − 3xy 3 y 0 − x3 y 0 = y 3 + 3x2 y − 4x(y + x)3
y 0 (4(y + x)3 − 3xy 3 − x3 ) = y 3 + 3x2 y − 4x(y + x)3
y 3 + 3x2 y − 4x(y + x)3
y0 =
4(y + x)3 − 3xy 3 − x3
17.
y = ln
x2 + 3
√
x + ex
Answer:
y = ln(x2 + 3) −
y0 =
18. log3 (y) = sin(x) cos
Answer:
1
ln(x + ex )
2
2x
1 + ex
−
x2 + 3 2(x + ex )
x
y
9
x 0
)
(log3 (y)) = (sin(x) cos
y
y0
x
x 1
x 0
= cos(x) cos
− sin(x) sin( )
− y
ln(3)y
y
y
y y2
y0
x 0
x
x
− 2 y sin(x) sin( ) = cos(x) cos
− sin(x) sin(x/y) y −1
ln(3)y y
y
y
1
x
x
x
y0
− 2 sin(x) sin( ) = cos(x) cos
− sin(x) sin(x/y) y −1
ln(3)y y
y
y
cos(x) cos xy − sin(x) sin(x/y) (y −1 )
0
y =
1
− xy −2 sin(x) sin(x/y)
ln(3)y
0
19. tan(xy) =
Answer:
x
x3 +y 4 +xy
x
)0
+ y 4 + xy
(x3 + y 4 + xy) − x(3x2 + 4y 3 y 0 + y + xy 0 )
(x3 + y 4 + xy)2
(x3 + y 4 + xy − 3x3 − xy) − (4xy 3 y 0 + x2 y 0 )
(x3 + y 4 + xy)2
(x3 + y 4 + xy − 3x3 − xy) (4xy 3 y 0 + x2 y 0 )
− 3
(x3 + y 4 + xy)2
(x + y 4 + xy)2
(x3 + y 4 + xy − 3x3 − xy)
− sec2 (xy)y
(x3 + y 4 + xy)2
(x3 + y 4 + xy − 3x3 − xy)
− sec2 (xy)y
(x3 + y 4 + xy)2
(tan(xy))0 = (
sec2 (xy)(y + xy 0 ) =
sec2 (xy)y + sec2 (xy)xy 0 =
sec2 (xy)y + sec2 (xy)xy 0 =
(4xy 3 y 0 + x2 y 0 )
sec2 (xy)xy 0 + 3
=
(x + y 4 + xy)2
(4xy 3 + x2 )
0
2
y sec (xy)x + 3
=
(x + y 4 + xy)2
0
y =
10
x3
(x3 +y 4 +xy−3x3 −xy)
(x3 +y 4 +xy)2
sec2 (xy)x +
− sec2 (xy)y
(4xy 3 +x2 )
(x3 +y 4 +xy)2
√
20. y = (x + 2) x + 1 sin(x) cos(x) arctan(x)(x + 3)−2 e−5x
Answer:
1
ln(x + 1) + ln(sin(x)) + ln(cos(x)) + ln(tan−1 (x)) − 2 ln(x + 3) − 5x
2
1
= (ln(x + 2) + ln(x + 1) + ln(sin(x)) + ln(cos(x)) + ln(tan−1 (x)) − 2 ln(x + 3) − 5x
2
1
1
cos(x) sin(x)
1
2
=
+
+
−
+
−
−5
2
x + 2 2(x + 1) sin(x) cos(x) arctan(x)(x + 1) x + 3
1
1
cos(x) sin(x)
1
2
=
+
+
−
+
−
−5 y
x + 2 2(x + 1) sin(x) cos(x) arctan(x)(x2 + 1) x + 3
1
1
cos(x) sin(x)
1
2
=
+
+
−
+
−
−5 ·
x + 2 2(x + 1) sin(x) cos(x) arctan(x)(x2 + 1) x + 3
√
(x + 2) x + 1 sin(x) cos(x) arctan(x)(x + 3)−2 e−5x
ln(y) = ln(x + 2) +
(ln(y))0
y0
y
y0
y0
21. etan(ax+by) = sin(cx) + cos(dy)
Answer:
etan(ax+by)
(etan(ax+by) )0 = (sin(cx) + cos(dy))0
sec2 (ax + by)(a + by 0 ) = c cos(cx) − d sin(dy)y 0
y 0 b sec2 (ax + by)etan(ax+by) + d sin(dy)y 0 = c cos(cx) − a sec2 (ax + by)etan(ax+by)
y 0 (b sec2 (ax + by)etan(ax+by) + d sin(dy)) = c cos(cx) − a sec2 (ax + by)etan(ax+by)
y0 =
22. y x = cos(2arcsin(x) )
Answer:
11
c cos(cx) − a sec2 (ax + by)etan(ax+by)
(b sec2 (ax + by)etan(ax+by) + d sin(dy))
ln(y x ) = ln(cos(2arcsin(x) ))
(x ln(y))0 = (ln(cos(2arcsin(x) )))0
− sin(2arctan(x) ) ln(2)2arctan(x)
y0
ln(y) + x =
y
cos(2arcsin(x) ))
− sin(2arctan(x) ) ln(2)2arctan(x)
y0
x =
y
cos(2arcsin(x) ))
y0 =
y
x
1
1+x2
1
1+x2
− sin(2arctan(x) ) ln(2)2arctan(x)
cos(2arcsin(x) ))
− ln(y)
1
1+x2
!
− ln(y)
√
23. 3y cos(y sin(x)) = tan−1 ( x + y + ex )
Answer:
√
(3y cos(y sin(x)) )0 = (tan−1 ( x + y + ex ))0
ln(3)3y cos(y sin(x)) (y 0 cos(y sin(x)) + y(− sin(y sin(x))(y 0 sin(x) + y cos(x))) =
1 + y 0 + ex
1
√
√
=
1 + ( x + y + ex )2 2 x + y + ex
ln(3)3y cos(y sin(x)) (y 0 cos(y sin(x)) − y sin(y sin(x))y 0 sin(x)
1
1 + y 0 + ex
√
√
−y 2 sin(y sin(x)) cos(x) =
1 + ( x + y + ex )2 2 x + y + ex
y0
1
√
ln(3)3y cos(y sin(x)) (y 0 cos(y sin(x)) − y sin(y sin(x))y 0 sin(x) −
1 + x + y + ex 2 x + y + ex
1
1 + +ex
√
= y 2 sin(y sin(x)) cos(x)
1 + x + y + ex 2 x + y + ex
1
√
y 0 (ln(3)3y cos(y sin(x)) cos(y sin(x)) − y sin(y sin(x)) sin(x) −
)
x
(1 + x + y + e )(2 x + y + ex )
1
1 + +ex
√
= y 2 sin(y sin(x)) cos(x)
1 + x + y + ex 2 x + y + ex
x
y0 =
1
√1++e
y 2 sin(y sin(x)) cos(x) 1+x+y+e
x 2 x+y+ex
ln(3)3y cos(y sin(x)) cos(y sin(x)) − y sin(y sin(x)) sin(x) −
12
1 √
(1+x+y+ex )(2 x+y+ex )
24. y = xy Answer:
ln(y) = ln(xy )
ln(y) = y ln(x)
(ln(y))0 = (y ln(x))0
y0
y
= y 0 ln(x) +
y
x
0
y
y
− y 0 ln(x) =
y
x
1
y
y0
− ln(x) =
y
x
y/x
y 0 = −1
y − ln(x)
y2
y0 =
x − xy ln(x)
II. Find an equation for the tangent line to the curve at the given point.
1. y 2 + x2 = 2xy, (1, 1)
Answer:
(y 2 + x2 )0 = (2xy)0
2yy 0 + 2x = 2y + 2xy 0
2yy 0 − 2xy 0 = 2y − 2x
y 0 (2y − 2x) = 2y − 2x
2y − 2x
y0 =
2y − 2x
0
y =1
The tangent line is:
(y − 1) = (x − 1) =⇒ y = x
13
2. y 3 = x + sin(x + π2 ), (0, 1)
Answer:
π
(y 3 )0 = (x + sin(x + ))0
2
π
2 0
3y y = 1 + cos(x + )
2
π
)
1
+
cos(x
+
0
2
y =
3y 2
y 0 (0, 1) =
1 + cos( π2 )
3(12 )
1
=
3
The tangent line is:
1
(y − 1) = (x − 0)
3
3. y = ln(cos(x)), (0, 0)
Answer:
y0 =
y 0 (0, 1) =
− sin(x)
cos(x)
− sin(0)
cos(0)
=0
The tangent line is:
y=0
x
4. y = x(x ) , (2, 16)
Answer:
14
x
ln(y) = ln(x(x ) )
ln(y) = (xx ) ln(x)
ln(ln(y)) = ln(xx ln(x))
ln(ln(y)) = ln(xx ) + ln(ln(x))
ln(ln(y)) = x ln(x) + ln(ln(x))
(ln(ln(y)))0 = (x ln(x) + ln(ln(x)))0
y0
1
= ln(x) + 1 +
ln(y)y
ln(x)x
1
0
y = ln(x) + 1 +
y ln(y)
ln(x)x
1
y = ln(2) + 1 +
16 ln(16)
ln(2)2
1
= ln(2) + 1 +
64 ln(2)
ln(2)2
= 64(ln(2))2 + 64 ln(2) + 32
0
The tangent line is:
(y − 16) = 64(ln(2))2 + 64 ln(2) + 32 (x − 2)
15