Question

Need help with implicit differentiation!

xy^2+yx^2=6

Answer 1

What do you mean by partial derivatives? Should I use the product rule?

Answer 2

this is an implicit differentiation

Answer 3

But this requires implicit differentiation, so you rather treat y as a function of x. I recommend to rewrite the equation this way.

Answer 4

so u need the product rule as well

Answer 5

\[\Large x y(x)^2+y(x)x^2=6 \]

Answer 6

Where do I take it from there?

Answer 7

\[2xy \frac{dy}{dx}+y^2+\frac{dy}{dx}x^2+2xy=0\]

Answer 8

that is what we call implicit differentiation...

Answer 9

this is not partial diff...

Answer 10

Can you please explain why you pu in the dy/dx?

Answer 11

because the question says find the implicit differentiation, which a fondamental subject for differential equations... now if you mean how did i get that result, well that is simple to see,,,

Answer 12

Ok I have \[P _{0}\] (1,2) also.

Where do I plug those in?

Answer 13

first of all xy^2+yx^2=6 as you can see x is multiplied by y^2 so you need to use the product rule, similarly you have yx^2 which requires product rule of differentiation...

Answer 14

next step you need to seperate variables...

Answer 15

what is Po? you need to provide more information, is it the tangent point, what is it?

Answer 16

Can I add 2xy\[\frac{ dy }{ dx }\] + 2xy or is thattwo completely different things.

The question says: Use the method of implicit differentiation to calculate dy/dx at the point Po

Answer 17

ok so it is the slope of the tangent, so now you need to solve for dy/dx, and yes you can add 2xy... like solving a normal algabraec problem.,..

Answer 18

that mean dy/dx should = something...

Answer 19

if you need help let me know, put dy/dx on one side and the rest on the other side...

Answer 20

or just plug in the numbers as it is, that would be easier for you

Answer 21

put for every x=1 and for every y=2

Answer 22

don't waste your time on rearranging the formula...

Answer 23

\[2(1)(2) \frac{dy}{dx}+(2)^2+\frac{dy}{dx}(1)^2+2(1)(2)=0\]

Answer 24

Okay, when I plug in the x and y to the formula, with out rearranging it, i don't plug it into dy/dx also do I?

Answer 25

Okay, thank you!

Answer 26

you're welcome!

Answer 27

ok so now i solve and leave the dy/dx in?

Answer 28

yes dy/dx is what you want to find at that point...

Answer 29

ok so after i solved I end up with 4 dy/dx + dy/dx +8.

How do I find dy/dx?

Answer 30

take dy/dx as your common factor

Answer 31

Sorry I still don't understand

Answer 32

Oh so 5 dy/dx + 8 =0?

Answer 33

It worked!! Thank you SO MUCH!

Answer 34

\[4\frac{dy}{dx}+\frac{dy}{dx}+8=0 \rightarrow 5\frac{dy}{dx}=-8\]

Answer 35

mr wislar stop those pictures please

Answer 36

thank you

Answer 37

Ok so at the beginning of the problem, how do I know when to incorporate in the dx/dy?

Answer 38

your final answer is dy/dx=-8/5

Answer 39

Yes I got that! Thanks! But now i'm moving on and I don't understand

how you got the equation 2xy dy/dx + y^2 + dy/dx x^2

Answer 40

+2xy

Answer 41

ok i will show you

Answer 42

consider the following example i want to avoid using termonologies ok...

Answer 43

ok

Answer 44

\[x^4+y^4=4\]

Answer 45

\[\frac{d(x^4)}{dx} + \frac{d(y^4)}{dx}=\frac{d(4)}{dx}\]

Answer 46

4x^3 is the first part do you agree with that?

Answer 47

diff x^4 =4x^3

Answer 48

diff(4)=0 do you agree?

Answer 49

now when you want to differentiat y^4 with respect to x, you get into the chain rule world

Answer 50

diff(y^4)=(4y^3)y'

Answer 51

ok

Answer 52

so the only new thing is when you differentiate the y?

Answer 53

we will report an abuse for any distraction ,,, stop sending pictures

Answer 54

we don't want viruses

Answer 55

Why did you multiply dy/dx to y, but add dy/dx to x?

Answer 56

sorry I mean y^2 and x^2

Answer 57

ok i will prove that for you it is based on the chain rule principle combined with the product rule...

Answer 58

the prodcut rule says you differentiate the first part leave the 2nd function , then it says leave the first part and differentiate the 2nd function

Answer 59

\[D(x^2y^2)= x^2 2yy'+2y^2x\]

Answer 60

did u get that?

Answer 61

Ok I think so

Answer 62

ok in a normal case how would diff xe^x?

Answer 63

Would we use the chain rule?

Answer 64

no the product rule...

Answer 65

when you have two functions muliplied by each other you use the product rule

Answer 66

Thank you so much for your patience and help! I have to go because my battery is dying. Your help is greatly appreciated and I did learn from you thank you so much again!

Answer 67

you're more than welcome! all the best