5 Steps to a 5 AP Calculus BC 2019

96 STEP 4. Review the Knowledge You Need to Score High

3. Findf′(x)iff(x)=ln(3x).

Answer: f′(x)=

1

3 x

(3)=

1

x

.

4. Find the approximate value of f′(3). (See Figure 6.7-1.)

y

x

f

(4,3)

(2,1)

0

Figure 6.7-1

Answer: Using the slope of the line segment joining (2, 1) and (4, 3),

f′(3)=

3 − 1

4 − 2

=1.

5. Find
   dy
   dx

ifxy= 5 x^2.

Answer: Using implicit differentiation, 1y+x

dy

dx

= 10 x. Thus,

dy

dx

=

10 x−y

x

.

Or simply solve foryleading toy= 5 xand thus,

dy

dx

= 5.

6. Ify=

5

x^2

, find

d^2 y

dx^2

.

Answer: Rewritey= 5 x−^2. Then,

dy

dx

=− 10 x−^3 and

d^2 y

dx^2

= 30 x−^4 =

30

x^4

.

7. Using a calculator, write an equation of the line tangent to the graphf(x)=
   − 2 x^4 at the point where f′(x)=−1.
   Answer: f′(x)=− 8 x^3. Using a calculator, enter [Solve](− 8 x∧ 3 =−1,x) and
   obtainx=

1

2

⇒ f′

(

1

2

)

=−1. Using the calculatorf

(

1

2

)

=−

1

8

. Thus,

tangent isy+

1

8

=− 1

(

x−

1

2

)

.

8. xlim→ 2
   x^2 +x− 6
   x^2 − 4
   Answer: Since

x^2 +x− 6

x^2 − 4

→

0

0

, consider limx→ 2

2 x+ 1

2 x

=

5

4

.

9. xlim→∞
   lnx
   x
   Answer: Since
   lnx
   x

→

∞

∞

, consider limx→∞

1 /x

1

=xlim→∞

1

x

=0.