Differentiation 99
- 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:
(
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 writey=
(2x+1)^1 /^2 (2x−1)^1 /^2 and use the product
rule.
- Letu= 2 x−1,
dy
dx
=10−csc^2 (2x−1)
=−20 csc^2 (2x−1). - 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)]
- Using the chain rule, letu= 2 x.
dy
dx
= 8
⎛
⎝√ −^1
1 −(2x)^2
⎞
⎠(2)=√−^16
1 − 4 x^2
- 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)
- Letu=(x^2 +3),
dy
dx
=
(
1
x^2 + 3
)
(2x)
=
2 x
x^2 + 3
.
Part B Calculators are allowed.
- 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