Question

please help solve
5 x^2+ 3 x + xy = 4 and y( 4 ) = -22, find y'( 4 ) by implicit differentiation.

Answer 1

5 x^2+ 3 x + xy = 4

10x+3 +x'y + xy' = 0

10x+3 +y + xy' = 0 solve for y' and plug in a 4

Answer 2

I know that you must make the equation y equal:
y=-(5x^2+3x-4)/x

Answer 3

implicit means dont make it equal to y ;)

Answer 4

I know but to sove it it would be easier to do that

Answer 5

The derivative I think would be:
(10x+3)/x^2

Answer 6

in this case, maybe; implicits are fairly easy. The only caveat is that you get a y in the implicit that would have to be subbed in from the explicit anyhows

Answer 7

Oh forget it I don't know what implicit means

Answer 8

lol, implicit means solve it as tho you solved for y; but without solving for y

Answer 9

I gotta learn how to do that

Answer 10

i was solving it explicit and but I just couldn't get how to work the y( 4 ) = -22 and the find y'( 4)

Answer 11

implicitly**

Answer 12

i got it, since y(4) = -22; just sub that in for y

10x+3 +y + xy' = 0

10(4)+3 +-22 + (4)y' = 0

40+3 -22+ (4)y' = 0

21+ (4)y' = 0

y' = -21/4

Answer 13

Oh that is cool

Answer 14

ahhh thank you so much, i think i was making it so much more complicated for myself and thank you for your help too rld613

Answer 15

implicit is just a name, and they really should teach you that its nothing new from what your used to

Answer 16

all you do is NOT throw out your derived bits; those x' and y'

Answer 17

the rules for the rest of it are exactly the same

Answer 18

product rule:

[fg]' = f'g+fg'

right? it doenst matter what the letters are:

[xy]' = x'y+xy'

Answer 19

since your used to deriving with respect to "x"; that x' (or rather dx/dx) = 1 and you toss it by habit

Answer 20

That is true

Answer 21

but ... you are perfectly content with y going to y' from the start

Answer 22

I have never done it like that before

Answer 23

tell me, what is the derivative of x^3?

Answer 24

3x^2

Answer 25

almost; i never stated with respect to x; so its actually 3x^2 x' until we KNOW what to do with the x'

Answer 26

whats the derivative of 2y^2?

Answer 27

4yy'

Answer 28

exactly

Answer 29

I don't understand why?

Answer 30

the rules and processes are not changed, you just learn not to toss out the derived bits

Answer 31

I don't understand why?

Answer 32

I don't understand why?

Answer 33

I don't understand why?

Answer 34

I don't understand why?

Answer 35

I don't understand why?

Answer 36

I don't understand why?

Answer 37

I don't understand why?

Answer 38

lets try it this manner:

\[\frac{d}{dx}x^3\]
\[\frac{dx}{dx}3x^2\]
\[1*3x^2\]
\[3x^2\]
right?

Answer 39

ya

Answer 40

that is what you are used to doing ... and you already do it for the y to begin with:
\[\frac{d}{dx}y\]
\[\frac{dy}{dx}1\]
\[\frac{dy}{dx}=y'\]

Answer 41

or maybe:
\[\frac{d}{dx}2y^2\]
\[\frac{dy}{dx}4y\]

Answer 42

y is it negative?

Answer 43

the posted problem is negative becasue that is the slope at x=4

Answer 44

it works out the same whether it was explicit or implicit

Answer 45

I gotta lookt his over. hopefully i will figure it out. It as to sink in

Answer 46

good luck with it :)

Answer 47

thanks for ur time