Document

Math 30 Homework #9
1. f (x) = x2 sin x
f 0 (x) = x2 cos x + (sin x)(2x) = 2x sin x + x2 cos x
(a) f 0 (x) = 2x sin x − x2 cos x
(c) f 0 (x) = 2x sin x + x2 cos x
(b) f 0 (x) = x2 sin x − 2x cos x
(d) f 0 (x) = −x2 sin x + 2x cos x
2. r(x) = (x3 − 7)5
r 0 (x) = 5(x3 − 7)4 (3x2 ) = 15x2 (x3 − 7)4
(a) r 0 (x) = 5x3 (x3 − 7)4
(b) r 0 (x) = 15x2 (x3 − 7)4
3. k(x) =
(c) r 0 (x) = 5x(x3 − 7)4
(d) r 0 (x) = (15x2 − 7)(x3 − 7)4
x cos x
e2x+3
d
d
e2x+3 dx
(x cos x) − (x cos x) dx
(e2x+3 )
(e2x+3 )2
e2x+3 [x(− sin x) + (cos x)(1)] − x cos x e2x+3 (2)
=
(e2x+3 )2
e2x+3 [−x sin x + cos x − 2x cos x]
=
(e2x+3 )2
cos x − x sin x − 2x cos x
=
e2x+3
k0 (x) =
(a) k0 (x) =
cos x + x sin x + 2x cos x
e2x+3
(b) k0 (x) =
cos x − x sin x + 2 cos x
e2x+3
4. m(x) =
p
cos x − x sin x − 2x cos x
e2x+3
cos x + x sin x − x cos x
(d) k0 (x) =
e2x+3
(c) k0 (x) =
sin(ex )
1
m(x) = (sin(ex)) 2
1 d
1
1
1
m0 (x) = (sin(ex ))− 2 (sin(ex )) = (sin(ex ))− 2 cos(ex )ex
2
dx
2
1
1
(sin(ex ))− 2 cos(ex )ex
2
1
1
(b) m0 (x) = (sin(ex ))− 2 sin(ex )
2
1
1
(c) m0 (x) = − (sin(ex ))− 2 cos(ex ) sin(ex )
2
(a) m0 (x) =
5. h(x) =
(d) m0 (x) =
1
1
(sin(ex ))− 2 sin(x)ex
2
1
sin (x + cos x)
2
h(x) =
1
= [sin(x + cos x)]−2
sin (x + cos x)
2
d
(sin(x + cos x))
dx
= −2 [sin(x + cos x)]−3 cos(x + cos x)(1 − sin x)
h0 (x) = −2 [sin(x + cos x)]−3
(a) h0 (x) = 2(sin(x + cos x))−3 sin(x + cos x)(1 − sin x)
(b) h0 (x) = 2(sin(x + cos x))−3 (1 − sin x)
(c) h0 (x) = −2(sin(x + cos x))−3 cos(x + cos x)(cos x)(1 − sin x)
(d) h0 (x) = −2(sin(x + cos x))−3 cos(x + cos x)(1 − sin x)
6. p(x) = sin
√
√
x − 1 · cos x + 2
√
√
1
1
p(x) = sin x − 1 · cos x + 2 = sin(x − 1) 2 cos(x + 2) 2
1 d
1
1
d cos(x + 2) 2 + cos(x + 2) 2
sin(x − 1) 2
dx
dx
1
1 d
1
1
1 d
1
= sin(x − 1) 2 − sin(x + 2) 2
(x + 2) 2 + cos(x + 2) 2 cos(x − 1) 2
(x − 1) 2
dx
dx
1
1
1
1
1
1
− 12
− 12
2
2
2
2
= − sin(x − 1) sin(x + 2)
(x + 2) (1) + cos(x + 2) cos(x − 1)
(x − 1) (1)
2
2
√
√
√
√
− sin x − 1 sin x + 2 cos x + 2 cos x − 1
√
√
=
+
2 x+2
2 x−1
1
p0 (x) = sin(x − 1) 2
√
√
√
√
sin x − 1 · sin x + 2 cos x − 1 · cos x + 2
√
√
(a) p (x) =
−
2 x−1
2 x+2
√
√
√
√
cos x − 1 · sin x + 2 sin x − 1 · cos x + 2
0
√
√
(b) p (x) =
−
2 x−1
2 x+2
√
√
√
√
sin x − 1 · cos x + 2 cos x − 1 · sin x + 2
0
√
√
(c) p (x) =
−
2 x−1
2 x+2
0
√
√
√
√
cos x − 1 · cos x + 2 sin x − 1 · sin x + 2
√
√
(d) p (x) =
−
2 x−1
2 x+2
0
cos x
7. g(x) = e
(b)
(c)
(d)
x+1
ex
x+1
x + 1 d cos x
g (x) = e
sin
+ sin
(e
)
ex
ex
dx
x+1
x+1 d x+1
+
sin
(ecos x )(− sin x)
= ecos x cos
ex
dx
ex
ex
x
x+1
e (1) − (x + 1)ex
x+1
cos x
cos x
=e
cos
−e
sin
sin x
ex
(ex )2
ex
x+1
−xex
x+1
cos x
−
e
sin
x
sin
= ecos x cos
ex
(ex)2
ex
x
x+1
x+1
cos x
cos x
−e
sin x sin
= −e
cos
ex
ex
ex
x+1
x+1 x
0
cos x
cos x
g (x) = −e
sin
+e
sin
ex
ex
ex
x+1
x+1 x
0
cos x
cos x
g (x) = −e
sin x · cos
−e
sin
ex
ex
ex
x+1
x+1 x
g 0 (x) = −ecos x sin x · sin
− ecos x cos
x
e
ex
ex
x+1
x+1 x
0
cos x
cos x
g (x) = −e
sin x · sin
−e
sin
ex
ex
ex
0
(a)
sin
cos x
d
dx
8. n(x) = sin(sin(cos x))
d
(sin(cos x))
dx
= cos(sin(cos x))(cos(cos x)(− sin x))
= − cos(sin(cos x)) cos(cos x) sin x
n0 (x) = cos(sin(cos x))
(a) n0 (x) = − cos(sin(cos x)) · cos(cos x) sin x
(b) n0 (x) = − cos(cos(sin x)) · sin x sin x
(c) n0 (x) = − cos(sin(cos x)) · sin(cos x) cos x
(d) n0 (x) = − cos(sin(cos x)) · sin x cos x
9. f (x) = x tan(1 + e2x )
d
tan(1 + e2x ) + tan(1 + e2x )(1)
dx
2
2x d
2x
= x sec (1 + e ) (1 + e ) + tan(1 + e2x )
dx
f 0 (x) = x
= x sec2 (1 + e2x)(e2x)(2) + tan(1 + e2x )
= 2x sec2 (1 + e2x )e2x + tan(1 + e2x)
(a) f 0 (x) = x tan(1 + e2x) + sec2 (1 + e2x)e2x
(b) f 0 (x) = tan(1 + e2x ) + 2x sec2 (1 + e2x )e2x
(c) f 0 (x) = tan(1 + e2x ) + sec2 (1 + e2x )e2x
(d) f 0 (x) = tan(1 + e2x ) + 2x sec2 (1 + e2x )
10. w(x) = sec(2 sin(πx))
d
(2 sin(πx))
dx
= sec(2 sin(πx)) tan(2 sin(πx))(2 cos(πx)(π))
w 0 (x) = sec(2 sin(πx)) tan(2 sin(πx))
= 2π sec(2 sin(πx) tan(2 sin(πx)) cos(πx)
(a) w 0 (x) = π sec(2 sin(πx)) tan(2 sin(πx)) sin(πx)
(b) w 0 (x) = 8πx sec(sin(πx)) tan(sin(πx)) cos(πx)
(c) w 0 (x) = 2πx sec(2 sin(πx)) tan(2 sin(πx)) sin(πx)
(d) w 0 (x) = 2π sec(2 sin(πx)) tan(2 sin(πx)) cos(πx)
11. v(x) =
√
√
csc x + sec x
1
1
v(x) = (csc x) 2 + sec x 2
1
1
1
1
v (x) = (csc x)− 2 (− csc x cot x) + (sec x 2 tan x 2 )
2
√
√
csc x cot x sec x tan x
√
=− √
+
2 csc x
2 x
0
1 −1
x 2
2
(a)
(b)
(c)
(d)
√
√
csc x cot x sec( x) tan( x)
√
+
v (x) = √
2 sec x
2 x
√
√
csc x cot x sec( x) tan( x)
0
√
√
v (x) =
+
2 x
2 sec x
√
√
csc x cot x sec( x) tan( x)
0
√
+
v (x) = − √
2 csc x
2 x
√
√
sec x tan x csc( x) cot( x)
0
√
v (x) = √
−
2 csc x
2 x
0
12. q(x) = cot2 (1 − x2 )
q(x) = cot2 (1 − x2 ) = cot(1 − x2 )
2
d
(cot(1 − x2 ))
dx
= 2 cot(1 − x2 )(− csc2 (1 − x2 ))(−2x)
q 0 (x) = 2 cot(1 − x2 )
= 4x cot(1 − x2 ) csc2 (1 − x2 )
(a) q 0 (x) = −2x cot(1 − x2 )(csc2 (1 − x2 ))
(c) q 0 (x) = 4x csc(1 − x2 )(cot2 (1 − x2 ))
(b) q 0 (x) = −2x2 cot(1 − x2 )(csc2 (1 − x2 ))
(d) q 0 (x) = 4x cot(1 − x2 )(csc2 (1 − x2 ))
13. h(x) = sin(sec2 x)
h(x) = sin(sec2 x) = sin((sec x)2 )
d
((sec x)2 )
dx
= cos(sec2 x)2 sec x(sec x tan x)
h0 (x) = cos((sec x)2 )
= 2 cos(sec2 x) sec2 x tan x
(a) h0 (x) = 2 cos(sec2 x) sec2 x tan x
(c) h0 (x) = cos(sec2 x) sec x tan x
(b) h0 (x) = 2 cos(sec2 x) sec x tan x
(d) h0 (x) = 2 cos(sec2 x) sin(sec2 x) tan x
14. m(x) = csc2 (csc x)
m(x) = csc2 (csc x) = (csc(csc x))2
d
(csc(csc x))
dx
= 2 csc(csc x)(− csc(csc x) cot(csc x))(− csc x cot x)
m0 (x) = 2(csc(csc x))
= 2 csc2 (csc x) cot(csc x) csc x cot x
(a) m0 (x) = 2 csc2 (csc x) csc x cot x
(c) m0 (x) = 2 csc2 (csc x) cot(csc x)
(b) m0 (x) = 2 csc2 (csc x) cot(csc x) csc x cot x (d) m0 (x) = −2 csc(csc x) cot(csc x) csc x cot x
15. Find an equation for the tangent line to g(x) = sin(cos(πx)) at x = 12 .
(a) y = −πx +
(b) y = πx −
π
2
π
2
(c) y =
1
2
(d) y = πx +
3π
2
We want to find the equation of the tangent line, so we need the slope and a point on the
line.
To find the slope we need the derivative.
d
(cos(πx))
dx
= cos(cos(πx))(− sin(πx))(π)
g 0 (x) = cos(cos(πx))
= −π cos(cos(πx)) sin(πx)
m = f0
1
1
1
= −π cos cos π ·
sin π ·
= −π(cos 0)(1) = −π(1) = −π
2
2
2
To find a point on the line,
at x = 21 . Therefore
we know the line touches the graph of g(x)
1
1
1
y = g( 2 ) = sin cos π · 2 = sin(0) = 0. Thus a point on the line is ( 2 , 0).
Now we can find the equation of the line.
1
y − 0 = −π x −
2
π
y = −πx +
2