practice with the chain rule & implicit differentiation

3
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 df dx for f HxL = 3 tanI2 x 2 +xMI-x 3 -x 2 M 12 x 5 -13 First, we rewrite our function in terms of “simpler” functions: f H xL = 3 tanI2 x 2 +xMI-x 3 -x 2 M 12 x 5 -13 = ΑH ΒH xLL ΓH xL ΔH xL = H H xL LH xL where ΑH xL = 3 tanH xL ΒH xL = I2 x 2 + xM ΓH xL = I- x 3 - x 2 M ΔH xL = I12 x 5 - 13M H H xL H ΒH xLL ΓH xL LH xL H xL which allows us to write f ' H xL = LH xL H ' H xL-H H xL L' H xL HLH xLL 2 = @ΔH xLD@Α' H ΒH xLL Β' H xLΓH xL' H xLΑH ΒH xLLD-@ΑH ΒH xLL ΓH xLD@Δ' H xLLD HΔH xLL 2

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Page 1: Practice with the Chain Rule & Implicit Differentiation

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

Page 2: Practice with the Chain Rule & Implicit Differentiation

. . . 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

Page 3: Practice with the Chain Rule & Implicit Differentiation

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