practice with the chain rule & implicit differentiation
TRANSCRIPT
Official SECTION 10 Document
Chain Rule & Implicit Differentiation
Let’s work on organizing our thoughts and computations to that, come Thurs-day, we may all stand victorious on a burning heap of derivatives.
1. Find d fd x
for f HxL =3 tanI2 x2+xM I-x3-x2M
12 x5-13
First, we rewrite our function in terms of “simpler” functions:
f HxL =3 tanI2 x2+xM I-x3-x2M
12 x5-13
=ΑHΒHxLL ΓHxL
∆HxL
=H HxLLHxL
where
ΑHxL = 3 tanHxLΒHxL = I2 x2 + xMΓHxL = I- x3 - x2M∆HxL = I12 x5 - 13MHHxL = ΑH ΒHxLL ΓHxLLHxL = ∆HxL
which allows us to write
f ' HxL =LHxL H ' HxL-H HxL L' HxL
HLHxLL2
=@∆HxLD@Α' HΒHxLL Β' HxL×ΓHxL+Γ' HxL×ΑHΒHxLLD-@ΑHΒHxLL ΓHxLD@∆' HxLLD
H∆HxLL2
. . . and, so we see that we should probably find Α ' H ΒHxLL, Β ' HxL, Γ ' HxL, and ∆ ' HxL.
Α ' HxL =d
d x3 tanHxL = 3 sec2HxL
Þ Α ' H ΒHxLL = 3 sec2H ΒHxLL = 3 sec2I2 x2 + xM
Β ' HxL =d
d xI2 x2 + xM = H4 x + 1L
(Notice that I kept parentheses around everything? This is so that I do not get confused when I plugΒ ' HxL back into my equation for f ' HxL.)
Γ ' HxL =d
d xI- x3 - x2M = I-3 x2 - 2 xM
∆ ' HxL =d
d xI12 x5 - 13M = I60 x4M
Now, we can simply (or rather, laboriously) plug these into our equation forf ' HxL which we have in terms of our simpler functions. Remember to keep all parenthe-
ses 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 x5-13EA3 sec2I2 x2+xM H4 x+1L I-x3-x2M+I-3 x2-2 xM 3 tanI2 x2+xME-A3 tanI2 x2+xM I-x3-x2MEA60 x4E
I12 x5-13M2
OK, so you might not get something this messy on the exam; but, if you canorganize this, you can do anything the test might throw at you.
2. Find d yd x
for H3 x y + 7L2 = 6 y
So, this looks like an implicit differentiation problem. In fact, it is.
H3 x y + 7L2 = 6 y
H1L Þd
d xAH3 x y + 7L2E =
dd x
@6 yDH2L Þ 2 H3 x y + 7L d
d x@3 x y + 7D = 6
d y
d x
H3L Þ 2 H3 x y + 7LB3 y +d y
d x3 xF = 6
d y
d x
H4L Þ 18 x y2 +d y
d x18 x2 y + 42 y +
d y
d x42 x =
d y
d x6
H5L Þd y
d xI18 x2 y + 42 x - 6M = -18 x y2 - 42 y
H6L Þd y
d x=
-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 dd x
AH3 x y + 7L2E, where 2 H3 x y + 7L is the derivative of the
“outside,” and dd x
@3 x y + 7D is the derivative of the “inside.”
(3) Used the product rule and implicit differentiation to find dd x
@3 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.
(5) We put all terms withd y
d x (that for which we are solving) on one side of the equation
sot hat we could factor out the d y
d x
(6) We divided both sides by I18 x2 y + 42 x - 6M in order to isolate d y
d x.
2 chain rule and implicit diff.nb
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 dd x
AH3 x y + 7L2E, where 2 H3 x y + 7L is the derivative of the
“outside,” and dd x
@3 x y + 7D is the derivative of the “inside.”
(3) Used the product rule and implicit differentiation to find dd x
@3 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.
(5) We put all terms withd y
d x (that for which we are solving) on one side of the equation
sot hat we could factor out the d y
d x
(6) We divided both sides by I18 x2 y + 42 x - 6M in order to isolate d y
d x.
Can you find y '' =d2 y
d x2 ?
chain rule and implicit diff.nb 3