Question

Find y' by implicit differentiation. e^(x/y)=6x−6y

Answer 1

I am really struggling with these problems. Can somebody walk me through it?

Answer 2

\[ \Large e^\frac{x}{y}=6x−6y\]????

Answer 3

Yes

Answer 4

taken logs?

Answer 5

\[e^{x/y}\frac{ y-x \frac{ dy }{ dx } }{ y^2 }=6-6\frac{ dy }{ dx}\]

Answer 6

I have, but I don't remember them to well.

Answer 7

\(\Large f (x,y) = e^\frac{x}{y}- 6x+ 6y = 0\)

\(\Large \nabla f = <\frac{1}{y} e^{\frac{x}{y}} - 6 , -\frac{x}{y^2} e^{x/y} +6>\)

\[\Large y' = -\dfrac{f_x}{f_y} = \dfrac{y e^{\frac{x}{y}} - 6y^2}{x e^{\frac{x}{y}} +6y^2} \]

Answer 8

soz, correction:
\[\Large \dfrac{dy}{dx} = -\dfrac{f_x}{f_y} = \color{red}{-}\dfrac{y e^{\frac{x}{y}} - 6y^2}{\color{red}{-}x e^{\frac{x}{y}} +6y^2} \\ \Large = \dfrac{6y^2 - y e^{\frac{x}{y}} }{6y^2-x e^{\frac{x}{y}} }\]

Answer 9

HI!!

Answer 10

that is one way to do it
you could do it without taking the log as well

Answer 11

the derivative of the right side is easy, it is \(6-6y'\)

Answer 12

the derivative of \[\frac{x}{y}\] is \[\frac{y-xy'}{y^2}\]

Answer 13

putting it together gives \[\frac{y-xy'}{y^2}e^{\frac{x}{y}}=6-6y'\]

Answer 14

@misty

Answer 15

Can you break it down more?

Answer 16

sure

Answer 17

which part is not clear? it is clear that taking the derivative of \[6x-6y\] with respect to \(x\) gives \[6-y'\]?

Answer 18

Yes. I am just having a hard time getting it to explicit form. y'=

Answer 19

oh that
that is a raft of algebra

Answer 20

\[\frac{y-xy'}{y^2}e^{\frac{x}{y}}-6y'=6-6y'\] maybe start with \[\frac{y-xy'}{y^2}e^{\frac{x}{y}}-6y'=6\]

Answer 21

break it up as \[\frac{y}{y^2}e^{\frac{x}{y}}-\frac{xyy'}{y^2}e^{\frac{x}{y}}-6y'=6\]

Answer 22

subtract and get \[-\frac{xy'}{y}-6y'=6-\frac{1}{y}e^{\frac{x}{y}}\]

Answer 23

factor out the \(y'\) on the left, then divide

Answer 24

oh i made a mistake in the algebra

Answer 25

\[-\frac{xy'}{y^2}-6y'=6-\frac{1}{y}e^{\frac{x}{y}}\]

Answer 26

I got it. Thanks!