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

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

Differentiation 133

6. Rewritey=

√

2 x+ 1

2 x− 1

as

y=

(

2 x+ 1

2 x− 1

) 1 / 2

. Applying first the chain
rule and then the quotient rule,
dy
dx

=

1

2

(

2 x+ 1

2 x− 1

)− 1 / 2

×

[

(2)(2x−1)−(2)(2x+1)

(2x−1)^2

]

=

1

2

1

(

2 x+ 1

2 x− 1

) 1 / 2

[

− 4

(2x−1)^2

]

=

1

2

1

(2x+1)^1 /^2

(2x−1)^1 /^2

[

− 4

(2x−1)^2

]

=

− 2

(2x+1)^1 /^2 (2x−1)^3 /^2

.

Note that

(

2 x+ 1

2 x− 1

) 1 / 2

=

(2x+1)^1 /^2

(2x−1)^1 /^2

,

if

2 x+ 1

2 x− 1

>0, which impliesx<−

1

2

orx>

1

2

.

An alternate method of solution is to write

y=

√

2 x+ 1

√

2 x− 1

and use the quotient rule.

Another method is to write

y=(2x+1)^1 /^2 (2x−1)^1 /^2 and use the

product rule.

7. Letu= 2 x−1,
   dy
   dx
   =10−csc^2 (2x−1)
   =−20 csc^2 (2x−1).


8. Using the product rule,
   dy
   dx

=(3[sec(3x)])+[sec(3x) tan(3x)](3)[3x]

=3 sec(3x)+ 9 xsec(3x) tan(3x)

=3 sec(3x)[1+ 3 xtan(3x)].

9. Using the chain rule, letu=sin(x^2 −4).

dy

dx

=10(−sin[sin(x^2 −4)])[cos(x^2 −4)](2x)

=− 20 xcos(x^2 −4) sin[sin(x^2 −4)]

10. Using the chain rule, letu= 2 x.

dy

dx

= 8

⎛

⎝√ −^1

1 −(2x)^2

⎞

⎠(2)=√−^16

1 − 4 x^2

11. Since 3e^5 is a constant, its derivative is 0.

dy

dx

= 0 +(4)(ex)+(ex)(4x)

= 4 ex+ 4 xex= 4 ex(1+x)

12. Letu=(x^2 +3),
    dy
    dx

=

(

1

x^2 + 3

)

(2x)

=

2 x

x^2 + 3

.

Part B Calculators are allowed.

13. Using implicit differentiation, differentiate
    each term with respect tox.

2 x+ 3 y^2

dy

dx

= 0 −

[

(5)(y)+

dy

dx

(5x)

]

2 x+ 3 y^2

dy

dx

=− 5 y− 5 x

dy

dx

3 y^2

dy

dx

+ 5 x

dy

dx

=− 5 y− 2 x

dy

dx

=(3y^2 + 5 x)=− 5 y− 2 x

dy

dx

=

− 5 y− 2 x

3 y^2 + 5 x

or

dy

dx

=

−(2x+ 5 y)

5 x+ 3 y^2

14. Sincef′(4) is equivalent to the slope of the
    tangent to f(x)atx=4, there are several
    ways you can find its approximate value.