MAT 16A
Homework 6
Finding derivatives:
1. f (x) =
sin x
.
x
Using quotient rule we obtain
d
d
[sin x] − sin x dx
[x]
x dx
2
sin x
x cos x − sin x
.
=
sin2 x
f 0 (x) =
2. f (x) = x2 sec x. Using product rule we have
d 2
d
[x ] sec x + x2 [sec x]
dx
dx
= 2x sec x + x2 sec x tan x
= x sec x(2 + x tan x).
f 0 (x) =
3. f (x) = sin x tan x. Using product rule we have
f 0 (x) =
=
=
=
=
d
d
[sin x] tan x + sin x [tan x]
dx
dx
2
cos x tan x + sin x sec x
sin x
cos x
+ sin x sec2 x
cos x
sin x + sin x sec2 x
(1 + sec2 x) sin x.
4. f (x) = ex sec x. Using product rule we have
d x
d
[e ] sec x + ex [sec x]
dx
dx
= ex sec x + ex sec x tan x
= ex (1 + tan x) sec x.
f 0 (x) =
5. f (x) =
cot x
.
1+x2
Using quotient rule we have
d
d
(1 + x2 ) dx
[cot x] − cot x dx
[1 + x2 ]
f (x) =
(1 + x2 )2
−(1 + x2 ) csc2 x − (cot x)(0 + 2x)
=
(1 + x2 )2
(1 + x2 ) csc2 x + 2x cot x
= −
.
(1 + x2 )2
0
1
6. f (x) = xex csc x. Using general product rule we have
d
d
d
[x]ex csc x + x [ex ] csc x + xex [csc x]
dx
dx
dx
= ex csc x + xex csc x − xex csc x cot x
= ex (1 + x − x cot x) csc x.
f 0 (x) =
Equation of tangent line:
1. f (x) = ex sin x. Using the product rule we have
d x
d
[e ] sin x + ex [sin x]
dx
dx
= ex sin x + ex cos x
= ex (sin x + cos x).
f 0 (x) =
Slope of the tangent line at x = 0 is f 0 (0) = e0 (sin 0 + cos 0) = 1. Also notice that
f (0) = e0 sin 0 = 0. Therefore equation of the tangent line at x = 0 is
i.e.,
y − f (0) = f 0 (0)(x − 0)
y = x.
2. f (t) = (1 + t2 ) cos t. Using product rule we have
d
d
[1 + t2 ] cos t + (1 + t2 ) [cos t]
dt
dt
= 2t cos t − (1 + t2 ) sin t.
f 0 (t) =
Therefore slope of the tangent line at x =
f
Notice that f
is
π
2
= 1+
0
π 2
π2
4
is
π
π
π2
π
= 2 cos − 1 +
sin
2
2
4
2
2
π
= − 1+
.
4
cos π2 = 0. Therefore equation of the tangent line at x =
π π
= f0
x−
2
2
2
π2 π
y =− 1+
x−
.
4
2
y−f
i.e.,
π
2
π 2
π
2
3. g(x) = e2 tan x. Since e2 is just a constant, it
to see that g 0 (x) = e2 sec2 x. Therefore
is easy
π
0 π
2
slope of the tangent line
at x = 4 is f 4 = e sec2 π4 = 2e2 . Therefore equation of the
tangent line at π4 , e2 is
π
2
2
y − e = 2e x −
4π 2
2
= 0.
i.e.,
y − 2e x − e 1 −
2
4. f (x) = xex sec x. Using the general product rule we have
d
d
d
[x]ex sec x + x [ex ] sec x + xex [sec x]
dx
dx
dx
= ex sec x + xex sec x + xex sec x tan x
= ex (1 + x + x tan x) sec x.
f 0 (x) =
Therefore slope of the tangent line at (0, 0) is
f 0 (0) = e0 (1 + 0 + 0 tan 0) sec 0
= 1.
Hence equation of the tangent line at (0, 0) is
i.e,
y − 0 = (x − 0)
y − x = 0.
3