Question

Using Implicit Differentiation, take the derivative of
d(x^2+yx^2+2(x^2)y+x^3)/dx

Answer 1

\[\frac{ d(x^2+yx^2+2x^2y+x^3) }{ dx }\]

Answer 2

\[d(x^2 +yx^2 +2(x^2)y+x^3)\]

Answer 3

Just take the derivative like you normally do, but wherever you see a y, take the derivative of the y function and tack on a y'

Answer 4

okay...

Answer 5

So what is the derivative of \(x^2\)?

Answer 6

2x

Answer 7

\(yx^2\)? Use product rule.

Answer 8

d(x^2)/dx y + dy/dx(x^2)

Answer 9

2xy + dy/dx(x^2)

Answer 10

Why do you write it like that, haha. I would simply write it as\[d(yx^2) = x^2y' +2xy\]

Answer 11

Yeah, same thing. Alright, good job \(\checkmark\)

Answer 12

mkay

Answer 13

Same thing with \(2x^2y\)

Answer 14

4xy + (2x^2)y(prime)

Answer 15

Oh and one more thing, is your function \(f(x) = x^2+yx^2+2(x^2)y+x^3\)?

Answer 16

yes

Answer 17

Okay, so far we have \[y' = 2x + x^2y'+2xy + 4xy+2x^2y'\]

Answer 18

Correct?

Answer 19

i believe so

Answer 20

Alright, now the derivative of \(x^3\)?

Answer 21

3x^2

Answer 22

And that finishes the derivative part.

Answer 23

SO we have \[y'= 2x + x^2y'+2xy + 4xy+2x^2y'+3x^2\]

Answer 24

yes

Answer 25

Now take everything that does not have a \(y'\) and subtract it so it's on the other side.

Answer 26

Take your \(y'\) on the left side, and subtract it so its on the right with all the rest of its friends.

Answer 27

Wait I think I get it.. you just factor out the d/dx right ?

Answer 28

yep.

Answer 29

alright thank-you... ill take from here

Answer 30

that's why I like writing it out as \(y'\) instead of \(\dfrac{dy}{dx}\)

Answer 31

Makes it look much prettier :)

Answer 32

THANK YOU~

Answer 33

No problem. Gl. happy New Year :O)