MA 3972-MA-Book May 9, 2018 10:9
Differentiation 135
d^2 y
dx^2
∣∣
∣∣
x=π/ 2
=4 cos
(
π
2
)
− 2
(
π
2
)(
sin
(
π
2
))
= 0 − 2
(
π
2
)
(1)=−π
Or, using a calculator, enter
d(2x−sin(x),x,2)x=
π
2
and obtain−π.
- Entery 1 =(x−1)^2 /^3 +2 in your calculator.
The graph ofy1 forms a cusp atx=1.
Therefore,f is not differentiable atx=1. - Differentiate with respect tox:
(1) cosy+
[
(−siny)
dy
dx
]
(x)= 0
cosy−xsiny
dy
dx
= 0
dy
dx
=
cosy
xsiny
dy
dx
∣∣
∣∣
x=2,y=π/ 3
=
cos(π/ 3 )
(2) sin(π/ 3 )
=
1 / 2
2
(√
3 / 2
)=
1
2
√
3
.
Thus, the slope of the tangent to the curve
at (2,π/3) ism=
1
2
√
3
. The slope of the
normal line to the curve at (2,π/3) is
m=−
2
√
3
1
=− 2
√
- Therefore, an
equation of the normal line is
y−π/ 3 =− 2
√
3(x−2).
- limx→ 3
x^2 − 3 x
x^2 − 9
=limx→ 3
2 x− 3
2 x
=
1
2
- limx→ 0 +
ln(x+1)
√
x
=xlim→ 0 +
1 /(x+1)
1 /(2
√
x)
=xlim→ 0 +
2
√
x
x+ 1
= 0
- limx→ 0
ex− 1
tan 2x
=limx→ 0
ex
2 sec^22 x
=
1
2
- limx→ 0
cos(x)− 1
cos(2x)− 1
=limx→ 0
−sinx
−2 sin(2x)
=limx→ 0
−cosx
−4 cos(2x)
=
1
4
- limx→∞
5 x+2lnx
x+3lnx
=xlim→∞
5 +(2/x)
1 +(3/x)
= 5
7.12 Solutions to Cumulative Review Problems
- The expression
lim
h→ 0
sin
(
π
2
+h
)
−sin
(
π
2
)
h
is
the derivative of sinxatx=π/2, which is
the slope of the tangent to sinxat
x=π/2. The tangent to sinxatx=π/2is
parallel to thex-axis.
Therefore, the slope is 0, i.e.,
limh→ 0
sin
(
π
2
+h
)
−sin
(
π
2
)
h
=0.
An alternate method is to expand
sin
(
π
2
+h
)
as
sin
(
π
2
)
cosh+cos
(
π
2
)
sinh.
Thus, limh→ 0
sin
(
π
2
+h
)
−sin
(
π
2
)
h
=
lim
h→ 0
sin
(
π
2
)
cosh+cos
(
π
2
)
sinh−sin
(
π
2
)
h