Question

Find (dy/dx) of √(xy)=x+3y help please!

Answer 1

start with
\[\frac{1}{2\sqrt{xy}}\times \left(y+xy'\right)=1+3y'\] and solve for \(y'\)

Answer 2

Why would you start with that? I don't understand how you changed it to that.

Answer 3

you are looking for the derivative with respect to \(x\) right? i.e. \(\frac{dy}{dx}\)

Answer 4

That is correct.

Answer 5

and the initial equation is, if i read it correctly
\[\sqrt{xy}=x+3y\]

Answer 6

That is also correct.

Answer 7

ok then we are thinking that \(y\) is a function of \(x\) even though we don't know a formula for it, i.e. \(y=f(x)\)
that is why it is called "implicit", because there is no explicit formula for \(f(x)\)
so lets change all the \(y\)'s to \(f(x)\)

Answer 8

Alright, I'm following...

Answer 9

\[\sqrt{xy}=x+3y\]
\[\sqrt{xf(x)}=x+3f(x)\]

Answer 10

the derivative of the right side is easy, it is
\[1+3f'(x)\] which i wrote more succinctly as
\[1+3y'\] so far so good?

Answer 11

Ok yes, I understand that now, Thanks. Continue,.

Answer 12

the left side is the one that is a pain
for that you need the product rule and the chain rule
before we do it, suppose you had to take the derivative of
\[\sqrt{x\sin(x)}\] would that be okay?

Answer 13

Yes it would!

Answer 14

ok you would get
\[\frac{1}{2\sqrt{x\sin(x)}}\times (\sin(x)+x\cos(x))\] by the chain rule and the product rule

Answer 15

okk

Answer 16

similarly for
\[\sqrt{xf(x)}\] you get the derivative as
\[\frac{1}{2\sqrt{xf(x)}}\times (f(x)+xf'(x))\]

Answer 17

replacing \(f(x)\) by \(y\) and \(f'(x)\) by \(y'\) you get \[\frac{1}{2\sqrt{xy}}\times (y+xy')\]

Answer 18

Ohhh ok

Answer 19

and then, like you said, solve for y'?

Answer 20

yes

Answer 21

the algebra is going to be a drag, but it is algebra

Answer 22

Thats not a problem! I'm pretty quick with Algebra.

Answer 23

have fun
hope it is more or less clear