86 STEP 4. Review the Knowledge You Need to Score High
Example 1y=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 havedy
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− 3du
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.