Chained 1011818

Suppose g is differentiable at x and f is differentiable of kN .

Then fog is differentiable at x ant

tog) IN -- f 'll 9TH .Intuitiveness : Write y .- flew and u -- geo . Then in Leibniz 's rotation . the chain role is :

Hittite .It is tempting to just cancel the du 's . but

' '

II is just a symbol for something we do to afunction of u handy differentiation)

. It is not a real fraction

,so cancelling it doesn't make sense .

The book gives a real proof in Section 3.4 .

Erland find steer .Solution refl is a composition of two functions .

'

If we let few be the"

inner function"

FINE x 't I

at flu be the " outer function"

flu =P,

Then footwear . So wecan use the chain

ruler. We just need to know f

' and d'

. Then

we can plug everything inpower rule lower and sun rules

free fare # emit he- Hifi fruitful E2x

Therefore,

tyre - Hod ' N

Phe in ourform was

= fibro . girl

for fig . and o'FEET. 2x=L

Nfl.

temple. find ¥ sink .There are fatleast two ways we could do this : Product rule or chain role .

If we use the product rule , he can compute :

¥ Sliestfine

. sink )

-

- thesheIsmet sine #sine )= cousin et sine cost

= 25in# se.

Let's check that the chain rule agrees with this answer . We apply it with N and sink

¥ sink - Hsing '

differentiate the 728inNo# sink )outer severe first=

zgnecosk.

temple find

Lacosse:You could still use the product rule

, in a

"

chain"

¥Coste= ¥Kosercos * cos= ¥cose) hose . cost ) those rose . rosel )=

- sine . cos 't those # cost dos et rosette cost )= - costs int t cost f Sinko set rose f sine ))

=- cosy sink - sink COSH - sine cosy

.

-- 3 cosy sink

.

Watch how much easier the chain rule is :

¥ cosh - - tf @setpower rule ?- 3 N ' cost )on therobe

=3 cost I - Sind

= - 3054 sing . ✓

Longview You can differentiate functions that look like

flysheet)

with the chain role,

or even longer chains . for this one . you get i

Fetherolf forwent Gernot

-

- f'll Well . give . tired .We used the chain rule twice

.

Example find Ie sinksAmrein.

Solution

t sin , karate cos Leoswww.ftueosltmkl/--eoskosHnWlif-sinKanellgtetmk

= - rosko skin sinHarrel . see 't .

Combining.'

Eiland find Tereus les - NVM

Solution ! first . use the product role , then use the chain role on each piece:

Just product role

Hunslet # Mit#knishes - run ' teens return )Now use the chain a snailrule on both derivatives=/ . 2) . It ' - et IftGetlls . 14h13 - xtll ?

(3×2-11)=1012tell 443 - til " t 413k - I fuel Is IN - yup .We can factor out a common

IN1/4/43 - HIP to simplify :

= 2. IS - tell t213eel Ike till Evil 'll ' - HIP .=L. Get the 't be ' -9k - 2) full MM - HIP

=L Has the . let 3) kettles - http .

Imagine doing this without the chain role !

team. find Ta '.

This time.

first do the chain role because the power is on the outside . Then to the quotient role .'

¥E¥i .- if filteril-

- off I'

.

KittieGuy

'

=

h's eius si combine this suite the ,.

4.

see-218

=

get'

Exponentials We can now compute ¥6 " for any boo .Remember

beefn

b

!Therefore . raising both sitesto the e givesb' yetnot=p

Knox

Thus we can apply the chain role with e' and until :

¥6 's tf elm blue

- unble dbecause skeeterIup .*

Hnbit

= elhb InH

-

- b'

Inlol .So

.for Hamp le

¥2 ' -- UH21 .

Snitched ! This formula for bee should give Feet . Indeed ,

¥ et - - et . Chef et , because he -1 .

thplicitdifferentiation Sometimes we are given on expression involving Kandy which is more

complicated in y than

f- f CN .

For instance . consider the equation for the circle ofradius Si

uterus.

There is no way to mate this the graph of a function becauseit fails the vertical line test

.

Yet

we can still make sense of II at any point .To to this

.Apply ¥ to both sides and use the doin rule on any expression involving ti

Hurt feedI I

It' t *

Y'

-

- O

2kt 24 . ¥ - O2e ¥ = - 2x .

if - eThis givesa relation between t , t and ¥ . But what does It mean if y is not a function of l?Well

,if I'm given any point call on my curve . then if I ignore part of my curve away

from call, what's left my be the graph of a function with the point 4,61 on it .

For eeamplo.

if fall M) on my circle . then I can ignore the bottom half of the circle where13.91 does not lie :

i i

to:-.#it .This " top half " is the graph of a function f because it passes the vertical line test . In fort ,five Ffa

.

and our relation I ¥ = - t is telling us that for my point law on the top half of the circle I forinstance

.for 174 ) ) we have : b. FIN = - a I in case fable out this gives

4ft③

=-3or five } . I