anonymous
  • anonymous
Find y' assuming that the equation determines a differentiable function f such that y= f(x) sin^2 3y=x+y-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
\(sin^2(3y)\) ??
anonymous
  • anonymous
its implicit differentiation
amistre64
  • amistre64
yeah; but whats the equation?

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anonymous
  • anonymous
yes amistre that is correct
anonymous
  • anonymous
its kind of assumed its sin^2 (3y) lol
anonymous
  • anonymous
its unlikely to be sin^2(3) y
amistre64
  • 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
  • amistre64
thats wrong ....
anonymous
  • anonymous
?
amistre64
  • amistre64
i did sin^2(3y) as tho it was simply sin(3y)
anonymous
  • anonymous
ohh yeh, I see
anonymous
  • anonymous
alot of chain rules in that question
anonymous
  • anonymous
its only once elecengineer
anonymous
  • anonymous
lol
amistre64
  • 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
  • amistre64
3 times actually
amistre64
  • amistre64
\[u^2.sin(v).3y\]
anonymous
  • anonymous
also, you couls apply double angle formula for sin to simplify the bottom :P
amistre64
  • amistre64
yup
anonymous
  • anonymous
im trying to figure out what you did amistre
anonymous
  • anonymous
consider the LHS first, the derivative of the RHS is easy as
anonymous
  • 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
  • anonymous
so d/dx [ (sin(3y) ) ^2 ] = 2 sin(3y) [d/dx ( sin(3y) ) ]
anonymous
  • anonymous
= 2 sin(3y) 3y' cos(3y)
amistre64
  • amistre64
say you wanna derive: u^2; thats simple right? but: u = sin(v) and v = 3y
anonymous
  • 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
  • anonymous
the derivative of 3y with respect to x is 3 (dy/dx) = 3y'
anonymous
  • anonymous
i get the 2(sin3y)(cos y) but not the 3y'
amistre64
  • 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
  • amistre64
you dont understand why we keep something hta tyou were taught to throw out
anonymous
  • anonymous
:S?
amistre64
  • amistre64
\[y = 2x^3 \rightarrow y' = 6x^2 x'\]
anonymous
  • anonymous
oh...
amistre64
  • amistre64
what did the \(y\) derive to? \(y'\) right?
amistre64
  • amistre64
3y derives to: \(3 y'\) \(5y^2\) derives to \(10y .y'\)
anonymous
  • anonymous
so the derivative of sin 3y is 3 sin 3y then 3y' cos 3y?
amistre64
  • amistre64
you are used to throwing out the x' bit; but that is only because \(dx\over dx\)=1
anonymous
  • anonymous
i know that ist the trig function thats throwing me off
amistre64
  • amistre64
sin(3y) derives to 3y' cos(3y)
anonymous
  • anonymous
yes...thats what confused me or should i say confuses
amistre64
  • amistre64
the sin to cos? or the innards to the outside?
anonymous
  • anonymous
the innards the 3y inside the sin 3y
amistre64
  • 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
  • amistre64
the 'chain' rule is just accounting for all the inputs and outputs thru the equation that eah have a controling part
anonymous
  • anonymous
ok i guess i have to practice some more examples of them
amistre64
  • amistre64
try it on simple ones first: like, compound functions
amistre64
  • amistre64
interactmath.com can help; its a free practice math site
anonymous
  • anonymous
THANKS AGAIN SO MUCH AMISTRE AND ELECENGINEER
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
can you help me with the question i posted earlier today

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