Official SECTION 10 Document
Chain Rule & Implicit Differentiation
Let’s work on organizing our thoughts and computations to that, come Thursday, we may all stand victorious on a burning heap of derivatives.
1. Find
df
dx
for f HxL =
3 tanI2 x2 +xM I-x3 -x2 M
12 x5 -13
First, we rewrite our function in terms of “simpler” functions:
f HxL =
=
=
3 tanI2 x2 +xM I-x3 -x2 M
12 x5 -13
ΑHΒHxLL ΓHxL
∆HxL
H HxL
LHxL
where
ΑHxL = 3 tanHxL
ΒHxL = I2 x2 + xM
ΓHxL = I- x3 - x2 M
∆HxL = I12 x5 - 13M
HHxL = ΑHΒHxLL ΓHxL
LHxL = ∆HxL
which allows us to write
f ' HxL =
=
LHxL H' HxL-HHxL L' HxL
HLHxLL2
@∆HxLD@Α' HΒHxLL Β' HxL×ΓHxL+Γ' HxL×ΑHΒHxLLD-@ΑHΒHxLL ΓHxLD@∆' HxLLD
H∆HxLL2
2
chain rule and implicit diff.nb
. . . and, so we see that we should probably find Α' HΒHxLL, Β' HxL, Γ ' HxL, and ∆' HxL.
Α' HxL =
d
dx
3 tanHxL = 3 sec2 HxL
Þ Α' HΒHxLL = 3 sec2 HΒHxLL = 3 sec2 I2 x2 + xM
Β' HxL =
d
dx
I2 x2 + xM = H4 x + 1L
(Notice that I kept parentheses around everything? This is so that I do not get confused when I plug
Β ' HxL back into my equation for f ' HxL.)
Γ' HxL =
∆' HxL =
d
dx
d
dx
I- x3 - x2 M = I-3 x2 - 2 xM
I12 x5 - 13M = I60 x4 M
Now, we can simply (or rather, laboriously) plug these into our equation for
f ' HxL which we have in terms of our simpler functions. Remember to keep all parentheses and brackets, so that multiplication doesn’t go awry.
f ' HxL =
=
LHxL H' HxL-HHxL L' HxL
HLHxLL2
@∆HxLD@Α' HΒHxLL Β' HxL×ΓHxL+Γ' HxL×ΑHΒHxLLD-@ΑHΒHxLL ΓHxLD@∆' HxLLD
H∆HxLL2
A12 x -13EA3 sec I2 x +xM H4 x+1L I-x3 -x2 M+I-3 x2 -2 xM 3 tanI2 x2 +xME-A3 tanI2 x2 +xM I-x3 -x2 MEA60 x4 E
5
=
2
2
I12 x5 -13M
2
OK, so you might not get something this messy on the exam; but, if you can
organize this, you can do anything the test might throw at you.
2. Find
dy
dx
for H3 x y + 7L2 = 6 y
So, this looks like an implicit differentiation problem. In fact, it is.
H1L Þ
H3 x y + 7L2 = 6 y
d
AH3
dx
x y + 7L2 E =
H2L Þ 2 H3 x y + 7L
d
@3
dx
H3L Þ 2 H3 x y + 7LB3 y +
H4L Þ 18 x y2 +
H5L Þ
H6L Þ
dy
dx
dy
dx
d
@6
dx
yD
x y + 7D = 6
dy
dx
3 xF = 6
18 x2 y + 42 y +
dy
dx
dy
dx
dy
dx
42 x =
dy
dx
I18 x2 y + 42 x - 6M = -18 x y2 - 42 y
dy
dx
=
6
-18 x y2 -42 y
18 x2 y+42 x-6
In case you missed it, here’s a step-by-step breakdown of what we did:
(1) Differentiated each side with respect to x, treating y as a function of x.
(2) Used the chain rule on
“outside,” and
d
@3
dx
d
AH3
dx
x y + 7L2 E, where 2 H3 x y + 7L is the derivative of the
x y + 7D is the derivative of the “inside.”
In case you missed it, here’s a step-by-step breakdown of what we did:
chain rule and implicit diff.nb
3
(1) Differentiated each side with respect to x, treating y as a function of x.
(2) Used the chain rule on
“outside,” and
d
@3
dx
d
AH3
dx
x y + 7L2 E, where 2 H3 x y + 7L is the derivative of the
x y + 7D is the derivative of the “inside.”
(3) Used the product rule and implicit differentiation to find
d
@3
dx
x y + 7D. (We used
f HxL = 3 x as our “first” function of x and gHxL = y as our “second” function of x.)
(4) We foiled that junk.
dy
(5) We put all terms with d x (that for which we are solving) on one side of the equation
sot hat we could factor out the
dy
dx
(6) We divided both sides by I18 x2 y + 42 x - 6M in order to isolate
Can you find y '' =
d2 y
d x2
?
dy
.
dx