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

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.
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