5 Steps to a 5 AP Calculus BC 2019

Differentiation 101

18.

dy

dx

=(2)(sinx)+(cosx)(2x)=

2 sinx+ 2 xcosx

d^2 y

dx^2

=2 cosx+[(2)(cosx)+(−sinx)(2x)]

=2 cosx+2 cosx− 2 xsinx

=4 cosx− 2 xsinx

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) ism=−

2

√

3

1

=− 2

√

3.

Therefore, an equation of the normal line

isy−π/ 3 =− 2

√

3(x−2).

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

=

1

2

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

= 5

6.11 Solutions to Cumulative Review Problems

26. The expression

limh→ 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.