5 Steps to a 5 AP Calculus AB 2019 - William Ma

(Marvins-Underground-K-12) #1
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.


  1. 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.

  2. 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




  1. Therefore, an
    equation of the normal line is
    y−π/ 3 =− 2



3(x−2).


  1. limx→ 3
    x^2 − 3 x
    x^2 − 9
    =limx→ 3
    2 x− 3
    2 x


=


1


2



  1. limx→ 0 +
    ln(x+1)

    x
    =xlim→ 0 +
    1 /(x+1)
    1 /(2



x)

=xlim→ 0 +

2



x
x+ 1

= 0



  1. limx→ 0
    ex− 1
    tan 2x
    =limx→ 0
    ex
    2 sec^22 x


=


1


2



  1. limx→ 0
    cos(x)− 1
    cos(2x)− 1
    =limx→ 0
    −sinx
    −2 sin(2x)


=limx→ 0
−cosx
−4 cos(2x)

=


1


4



  1. limx→∞
    5 x+2lnx
    x+3lnx
    =xlim→∞
    5 +(2/x)
    1 +(3/x)


= 5


7.12 Solutions to Cumulative Review Problems



  1. 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
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