UnknownRandom
  • UnknownRandom
Find y' by implicit differentiation. e^(x/y)=6x−6y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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UnknownRandom
  • UnknownRandom
I am really struggling with these problems. Can somebody walk me through it?
IrishBoy123
  • IrishBoy123
\[ \Large e^\frac{x}{y}=6x−6y\]????
UnknownRandom
  • UnknownRandom
Yes

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IrishBoy123
  • IrishBoy123
taken logs?
UnknownRandom
  • UnknownRandom
\[e^{x/y}\frac{ y-x \frac{ dy }{ dx } }{ y^2 }=6-6\frac{ dy }{ dx}\]
UnknownRandom
  • UnknownRandom
I have, but I don't remember them to well.
IrishBoy123
  • IrishBoy123
\(\Large f (x,y) = e^\frac{x}{y}- 6x+ 6y = 0\) \(\Large \nabla f = <\frac{1}{y} e^{\frac{x}{y}} - 6 , -\frac{x}{y^2} e^{x/y} +6>\) \[\Large y' = -\dfrac{f_x}{f_y} = \dfrac{y e^{\frac{x}{y}} - 6y^2}{x e^{\frac{x}{y}} +6y^2} \]
IrishBoy123
  • IrishBoy123
soz, correction: \[\Large \dfrac{dy}{dx} = -\dfrac{f_x}{f_y} = \color{red}{-}\dfrac{y e^{\frac{x}{y}} - 6y^2}{\color{red}{-}x e^{\frac{x}{y}} +6y^2} \\ \Large = \dfrac{6y^2 - y e^{\frac{x}{y}} }{6y^2-x e^{\frac{x}{y}} }\]
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
that is one way to do it you could do it without taking the log as well
misty1212
  • misty1212
the derivative of the right side is easy, it is \(6-6y'\)
misty1212
  • misty1212
the derivative of \[\frac{x}{y}\] is \[\frac{y-xy'}{y^2}\]
misty1212
  • misty1212
putting it together gives \[\frac{y-xy'}{y^2}e^{\frac{x}{y}}=6-6y'\]
UnknownRandom
  • UnknownRandom
@misty
UnknownRandom
  • UnknownRandom
Can you break it down more?
misty1212
  • misty1212
sure
misty1212
  • misty1212
which part is not clear? it is clear that taking the derivative of \[6x-6y\] with respect to \(x\) gives \[6-y'\]?
UnknownRandom
  • UnknownRandom
Yes. I am just having a hard time getting it to explicit form. y'=
misty1212
  • misty1212
oh that that is a raft of algebra
misty1212
  • misty1212
\[\frac{y-xy'}{y^2}e^{\frac{x}{y}}-6y'=6-6y'\] maybe start with \[\frac{y-xy'}{y^2}e^{\frac{x}{y}}-6y'=6\]
misty1212
  • misty1212
break it up as \[\frac{y}{y^2}e^{\frac{x}{y}}-\frac{xyy'}{y^2}e^{\frac{x}{y}}-6y'=6\]
misty1212
  • misty1212
subtract and get \[-\frac{xy'}{y}-6y'=6-\frac{1}{y}e^{\frac{x}{y}}\]
misty1212
  • misty1212
factor out the \(y'\) on the left, then divide
misty1212
  • misty1212
oh i made a mistake in the algebra
misty1212
  • misty1212
\[-\frac{xy'}{y^2}-6y'=6-\frac{1}{y}e^{\frac{x}{y}}\]
UnknownRandom
  • UnknownRandom
I got it. Thanks!

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