Question

Find y' assuming that the equation determines a differentiable function f such that y= f(x)

sin^2 3y=x+y-1

Answer 1

\(sin^2(3y)\) ??

Answer 2

its implicit differentiation

Answer 3

yeah; but whats the equation?

Answer 4

yes amistre that is correct

Answer 5

its kind of assumed its sin^2 (3y) lol

Answer 6

its unlikely to be sin^2(3) y

Answer 7

\[3y' 2\cos(3y)=1+y'\]
\[6y'\cos(3y)-y'=1\]
\[y'(6\cos(3y)-1)=1\]
\[y' = \frac{1}{6\cos(3y)}\]

Answer 8

thats wrong ....

Answer 9

?

Answer 10

i did sin^2(3y) as tho it was simply sin(3y)

Answer 11

ohh yeh, I see

Answer 12

alot of chain rules in that question

Answer 13

its only once elecengineer

Answer 14

lol

Answer 15

\[3y'.2(\sin(3y)).cos(3y)=1+y'\]
\[y'.6\sin(3y)\cos(3y)-y'= 1\]
\[y'(6\sin(3y)\cos(3y)-1)= 1\]
\[y' = \frac{1}{6\sin(3y)\cos(3y)-1}\]
maybe

Answer 16

3 times actually

Answer 17

\[u^2.sin(v).3y\]

Answer 18

also, you couls apply double angle formula for sin to simplify the bottom :P

Answer 19

yup

Answer 20

im trying to figure out what you did amistre

Answer 21

consider the LHS first, the derivative of the RHS is easy as

Answer 22

so ( sin(3y) )^2

apply the chain rule , bring the power down in front , leave the inside of the bracket alone , reduce the power by one , then multiplt by the derivative of the inside

Answer 23

so d/dx [ (sin(3y) ) ^2 ] = 2 sin(3y) [d/dx ( sin(3y) ) ]

Answer 24

= 2 sin(3y) 3y' cos(3y)

Answer 25

say you wanna derive:

u^2; thats simple right?

but:

u = sin(v)

and

v = 3y

Answer 26

when you differentiate sin(3y) you take the derivative of the inside with respect to x, and then change the sin function to a cos function

Answer 27

the derivative of 3y with respect to x is 3 (dy/dx) = 3y'

Answer 28

i get the 2(sin3y)(cos y) but not the 3y'

Answer 29

\[{du\over dx}={du\over dv}{dv\over dy}{dy\over dx}\]

\[\frac{d(sin^2(3y))}{dx}=2sin(3y)*cos(3y)*3y'\]

Answer 30

you dont understand why we keep something hta tyou were taught to throw out

Answer 31

:S?

Answer 32

\[y = 2x^3 \rightarrow y' = 6x^2 x'\]

Answer 33

oh...

Answer 34

what did the \(y\) derive to? \(y'\) right?

Answer 35

3y derives to: \(3 y'\)
\(5y^2\) derives to \(10y .y'\)

Answer 36

so the derivative of sin 3y is 3 sin 3y then 3y' cos 3y?

Answer 37

you are used to throwing out the x' bit; but that is only because \(dx\over dx\)=1

Answer 38

i know that ist the trig function thats throwing me off

Answer 39

sin(3y) derives to 3y' cos(3y)

Answer 40

yes...thats what confused me or should i say confuses

Answer 41

the sin to cos? or the innards to the outside?

Answer 42

the innards the 3y inside the sin 3y

Answer 43

the innards have a controling part in the function and have to be accounted for.

They are the driving force that produces the output for the sin function

Answer 44

the 'chain' rule is just accounting for all the inputs and outputs thru the equation that eah have a controling part

Answer 45

ok i guess i have to practice some more examples of them

Answer 46

try it on simple ones first:

like, compound functions

Answer 47

interactmath.com can help; its a free practice math site

Answer 48

THANKS AGAIN SO MUCH AMISTRE AND ELECENGINEER

Answer 49

youre welcome :)

Answer 50

can you help me with the question i posted earlier today