Find y' assuming that the equation determines a differentiable function f such that y= f(x)
sin^2 3y=x+y-1

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- jamiebookeater

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- amistre64

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

- anonymous

its implicit differentiation

- amistre64

yeah; but whats the equation?

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## More answers

- anonymous

yes amistre that is correct

- anonymous

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

- anonymous

its unlikely to be sin^2(3) y

- amistre64

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

- amistre64

thats wrong ....

- anonymous

?

- amistre64

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

- anonymous

ohh yeh, I see

- anonymous

alot of chain rules in that question

- anonymous

its only once elecengineer

- anonymous

lol

- amistre64

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

- amistre64

3 times actually

- amistre64

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

- anonymous

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

- amistre64

yup

- anonymous

im trying to figure out what you did amistre

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

say you wanna derive:
u^2; thats simple right?
but:
u = sin(v)
and
v = 3y

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

:S?

- amistre64

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

- anonymous

oh...

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

i know that ist the trig function thats throwing me off

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

the innards the 3y inside the sin 3y

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

try it on simple ones first:
like, compound functions

- amistre64

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

- anonymous

THANKS AGAIN SO MUCH AMISTRE AND ELECENGINEER

- amistre64

youre welcome :)

- anonymous

can you help me with the question i posted earlier today

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