Question

if y^2 - 3xy = x-5 find dy/dx

Answer 1

Implicit differentiation and product rule.

Answer 2

?

Answer 3

Ok work through it step by step. What is the derivative of y^2?

Answer 4

2y?

Answer 5

Yeah. If you learned the chain rule then you might remember that you then multiply by the derivative of the inside so it'd be\[2y \frac{ dy }{ dx? } = 2y*y'\]
that's a y prime

Answer 6

okay

Answer 7

Are you going to try the next part? \[\frac{d}{dx}3xy\]

Answer 8

i don't understand how to do that...

Answer 9

Well remember that you can take constants out of the differentiation.\[\frac{d}{dx}3xy = 3\frac{d}{dx}xy\]
and then use the product rule

Answer 10

back off bud this is mine

Answer 11

y^2 - 3xy = x-5
Differentiate everything with respect to x, remember when you differentiate y though you end up it with it being multiplied by y prime (y').
For example you would make the derivative of y^2 into 2yy'.
\[\huge y^2 - 3xy = x- 5\]
To start off take the derivative of each value...
Remember to use the product rule for 3xy

Answer 12

Give it a try, and post your attempt .

Answer 13

@bibby are you fighting to answer questions?

Answer 14

Nah. Me and shamil "shamil98" martinez have some playful scuffles over math. It's all in good fun, good nature, and good love

Answer 15

i am still confused, i have never used this equation before

Answer 16

\[y^2 - 3xy = x-5\] treat it like \[f^2(x) - 3xf(x) = x-5\]

Answer 17

Take the derivative of every value separately
\[\huge \frac{ d }{ dx } y^2 = 2yy'\]
\[\huge \frac{ d }{ dx } -3xy = -3y -3xy'\]
\[\huge \frac{ d }{ dx } x = 1\]
\[\huge \frac{ d }{ dx } -5 = 0\]
Put everything together.
\[\huge 2yy' -3y - 3xy' = 1\]
Group the y' and solve for y'


\[\huge y'(2y-3x) = 3y+1\]
\[\huge y' = \frac{ 3y+1 }{ 2y-3x }\]

Answer 18

Do you understand how i got that?

Answer 19

i think so i will try with the next one and see

Answer 20

my boy @shamil98 delivers it again!

Answer 21

sorry i still don't understand

Answer 22

what don't you understand? ..

Answer 23

The multiplication rule goes something like u'v + uv'

Answer 24

This gives a nice explanation of implicit differentiation, i found it useful.
http://www.dummies.com/how-to/content/how-to-differentiate-implicitly0.html

Answer 25

Are you calling my lass a dummy

Answer 26

no, the website just has good information with a poorly named domain.

Answer 27

That's what I thought

Answer 28

adoptive fathers don't count

Answer 29

but i don't understand what to do with xy shouldn't they be nonexistant because x^0 and y^0 don't exist?

Answer 30

xy = x^1 * y^1 = xy..

when differentiating xy, you use the product rule:
\[\huge f'g + fg'\]

Answer 31

for y^2 +3x = xy -5 i got 3-y/x-2y

Answer 32

\[\huge y^2 + 3x = xy -5\]
\[\huge 2yy' + 3 = y + xy' \]
\[\huge y'(2y-x) = y-3\]
\[\huge y' = \frac{ y-3 }{ 2y-x }\]

Answer 33

\[2yy′+3=y+xy′\]
\[yy′ -xy'=\frac{ y-3 }{ 2 }\]
\[y'(y -x)=\frac{ y-3 }{ 2 }\]

shouldn't it be 2y - 2x? or 2(y-x)

Answer 34

2yy' + 3 = y+xy'

2yy' - xy' = y -3
y'(2y-x) = y-3
y' = y-3 / 2y - x
you can't just take the 2 out -_-

Answer 35

shut your mouth. Don't give me a moronic smiley because you skipped a step of factoring.

You've got this big fish in a little pond syndrome and it seriously gets under my skin brother

Answer 36

Admit that you are wrong, young one.
No shame in it , dear :)