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

MA 3972-MA-Book May 9, 2018 10:9

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

Example 1

y=e^3 x+ 5 xe^3 +e^3

dy

dx

=(e^3 x)(3)+ 5 e^3 + 0 = 3 e^3 x+ 5 e^3 (Note thate^3 is a constant.)

Example 2

y=xex−x^2 ex

Using the product rule for both terms, you have

dy

dx

=(1)ex+(ex)x−

[

(2x)ex+(ex)x^2

]

=ex+xex− 2 xex−x^2 ex=ex−xex−x^2 ex

=−x^2 ex−xex+ex=ex(−x^2 −x+1).

Example 3

y= 3 sinx

Letu=sinx. Then,

dy

dx

=(3sinx)(ln 3)

du

dx

=(3sinx)(ln 3) cosx=(ln 3)(3sinx) cosx.

Example 4

y=e(x^3 )

Letu=x^3. Then,

dy

dx

=

[

e(x^3 )

]du

dx

=

[

e(x^3 )

]

3 x^2 = 3 x^2 e(x^3 ).

Example 5

y=(lnx)^5

Letu=lnx. Then,

dy

dx

=5(lnx)^4

du

dx

=5(lnx)^4

(

1

x

)

=

5(lnx)^4

x

.

Example 6

y=ln(x^2 + 2 x−3)+ln 5

Letu=x^2 + 2 x−3. Then,

dy

dx

=

1

x^2 + 2 x− 3

du

dx

+ 0 =

1

x^2 + 2 x− 3

(2x+2)=

2 x+ 2

x^2 + 2 x− 3

.

(Note that ln 5 is a constant. Thus, the derivative of ln 5 is 0.)

Example 7

y= 2 xlnx+x

Using the product rule for the first term,

you have

dy

dx

=(2) lnx+

(

1

x

)

(2x)+ 1 =2lnx+ 2 + 1 =2lnx+3.