Question

Implicit differentiation

Answer 1

\[2(x^2+y^2)^2=25(x^2-y^2)\]

Answer 2

So far, I have:
\[4(X^2+y^2)(2x+2yy')=25(2x-2yy')\]

Answer 3

Cool :)

Answer 4

big X is a small one oops

Answer 5

You'll have to distribute some stuff.
I would recommend not FOIL'ing out the left side,\[\large\rm 4(x^2+y^2)(2x)+4(x^2+y^2)(2yy')\]Don't distribute all the way unless you see a need to.
Should make moving things around easier.

Answer 6

Oh maybe divide a 2 out of each side, ya?

Answer 7

as in, divide two to both sides to what I'm on at the moment?

Answer 8

Yes.\[\large\rm 4(x^2+y^2)(x+yy')=25(x-yy')\]

Answer 9

would I have to simplify \[4(x^2+y^2) \times (2x) \] first?

Answer 10

Did taking the 2's out confuse you?
The left side would look different if the 2's are gone.

Answer 11

yea I don't know how to get rid of the 2's inside the ( )

Answer 12

\[\large\rm 4(x^2+y^2)(2x+2yy')=25(2x-2yy')\]Le's factor a 2 out of each term in second set of brackets,\[\large\rm 4(x^2+y^2)\cdot2(x+yy')=25(2x-2yy')\]Dividing a 2 out of each term in the last set of brackets,\[\large\rm 4(x^2+y^2)\cdot2(x+yy')=25\cdot2(x-yy')\]Divide each side by 2,\[\large\rm 4(x^2+y^2)\cdot\cancel2(x+yy')=25\cdot\cancel2(x-yy')\]Ok with those steps? :)

Answer 13

Oh, I see it now.

Answer 14

Bahh, "divide out", "factor out", i switched between those wordings,
hopefully that wasn't too confusing.

Answer 15

\[\large\rm 4(x^2+y^2)(x+yy')=25(x-yy')\]Distributing is a little nicer now, no extra 2's floating around,\[\large\rm 4x(x^2+y^2)+4y(x^2+y^2)y'=25x-25yy'\]

Answer 16

Note that \(\large\rm 4(x^2+y^2)(yy')=4y(x^2+y^2)y'\)
It's just a habit that we usually write the primes in the back.

Answer 17

\[\large\rm \color{royalblue}{4x(x^2+y^2)}+\color{orangered}{4y(x^2+y^2)y'}=\color{royalblue}{25x}-\color{orangered}{25yy'}\]We would like to get all of our y' on the same side,
so in other words, blue on one side, orange on the other.

Answer 18

I am finishing the problem now

Answer 19

ok what I have so far is

Answer 20

\[4(x^2+y^2)(x+yy')=25x-25yy'\]

Answer 21

how would i tackle the left side

Answer 22

\[(4x^2+4y^2)(x+yy')\]?

Answer 23

Then foil?

Answer 24

\[\large\rm \color{orangered}{4(x^2+y^2)}(x+yy')=25x-25yy'\]Le sigh... if you want to :P yes
If this distributing is giving you a some trouble heh,\[\large\rm \color{orangered}{4(x^2+y^2)}x+\color{orangered}{4(x^2+y^2)}yy'=25x-25yy'\]

Answer 25

Okay I got it now fully. The distribution part got me, the distribution by taking out the common factor.

Answer 26

\[4x(x^2+y^2)+4yy'(x^2+y^2)=25x-25yy'\]
\[4yy'(x^2+y^2)+25yy'=25x-4x(x^2+y^2)\]
\[y'=\frac{ 25x-4x(x^2+y^2 }{ 4y(x^2+y^2)+25y }\]

Answer 27

Ooo looks great! :)

Answer 28

could u help me with another one? I'm like 85% done with it

Answer 29

sure