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

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.

19. 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.


20. 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

√

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

√

3(x−2).

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

=

1

2

22. limx→ 0 +
    ln(x+1)
    √
    x
    =xlim→ 0 +
    1 /(x+1)
    1 /(2

√

x)

=xlim→ 0 +

2

√

x

x+ 1

= 0

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

=

1

2

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

=limx→ 0

−cosx

−4 cos(2x)

=

1

4

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

= 5

7.12 Solutions to Cumulative Review Problems

26. 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