Question

Calculas

Answer 1

bring it :)

Answer 2

Find dy/dx by implicit Differentiation \[\sqrt{xy} = x-2y\]

Answer 3

square both sides then solve dy/dx of xy=x^2-4xy+4y^2

Answer 4

let me know if you need further help

Answer 5

so left side will equal 1y'=.....

Answer 6

How do I get dy/dx?

Answer 7

I tried using the chain rule but Im pretty sure Im doing it wrong.

Answer 8

you could do it this way:
as you should know, \(y'(x)=-\large\frac{F_x}{F_y}\)
where \(F(x,y)=x-2y-\sqrt{xy}\)
So: \(F_x=1-\large\frac{y}{\sqrt{xy}}\) and \(F_y=-2-\large\frac{x}{\sqrt{xy}}\)
now just find the quotient with oposite sign and you got your answer.

Answer 9

In previous post I forgot 2 in the denominator in expretions of F_x and F_y, as probably you noticed.
other way:
differentiating iplicitly:
\(\large\frac{xy'+y}{2\sqrt{xy}}=1-2y'\)
express y' from here to get the answer

Answer 10

So we're finding y prime right?

Answer 11

Yes sir..

Answer 12

\[y' = \frac{ xy^{2} }{-4-2x } - \frac{ 1 }{ -2-x }\]That's what I got.

Answer 13

Can you draw it please?

Answer 14

I think I need others to check it as well.

Answer 15

I'm wrong. I just saw an error.

Answer 16

Yeah, I dont think thats the answer. I tried Myko's solution, I still dont think I did it right.

Answer 17

Oh haha okay

Answer 18

But I will solve it correctly!

Answer 19

Its okay you really dont need to. I'm failing calculas anyway.

Answer 20

Lol.I remember I started doing well after the first midterm.

Answer 21

Are you serious? My AP exam is after the midterms

Answer 22

What's AP?

Answer 23

Its an exam we have to take if we are taking a college level class.

Answer 24

I think I got it right

Answer 25

\[y' = \frac{ 2\sqrt{xy}-y }{x+4\sqrt{xy}} \]