Question

How do I derive 2y^4 - x^4 = 3xy + 4? I get that I need to use implicit differentiation, but keep getting incorrect negative/positive values.

Answer 1

Wait is that your function and you have to find the derivative of it?

Answer 2

\[\huge f(x) = 2y ^{4}-x ^{4}=3xy+4\]

Answer 3

Yeah... it could be rewritten y = (2y^4 - x^4 - 4)/3x

Answer 4

ok just to make sure again, it's exactly what I wrote right?

Answer 5

I think so.

Answer 6

I trying to find dy/dx.

Answer 7

\[\huge \frac{ d }{ dx }(-x ^{4}+2^{4})=\frac{ d }{ dx }(4+3xy)\]

\[\huge 2\frac{ d }{ dx }(y^4)-\frac{ d }{ dx }(x^4) = \frac{ d }{ dx }(4+3xy)\]

Answer 8

Do the left side than do the right, do you see where this is going?

Answer 9

The first line should say 2y^4

Answer 10

Yep, I get \[8y^3 \frac{ dy }{ dx } -4x^3 = 3y + 3x \frac{ dy }{ dx }\]

Answer 11

Rearranging that gives \[\frac{ dy }{ dx } = \frac{ 3y = 4x^3 }{ 8y^3- 3x }\]

Answer 12

Opps, on top should be 3y + 4x^3

Answer 13

But it should be \[- \frac{ 3y-4x^3 }{ 3x-8y^3 }\]

Answer 14

Alright, sorry about that, I was helping someone else :), ok lets see where you went wrong.

Answer 15

3y+3x*dy/dx how did you get that? (right side)

Answer 16

Product rule. ie (3x)'y + 3x(y)'
Gives 3y + 3x dy/dx
I may have done that completely wrong.

Answer 17

Alright I'm just going to do the right side for you step by step.

Answer 18

\[\frac{ d }{ dx }(4)+3\frac{ d }{ dx }(xy)\]

\[3(\frac{ d }{ dx }(xy))+ 0\]

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

\[3(\frac{ dy }{ dx }(x)+(\frac{ d }{ dx }(x))y)\]

\[3y+3x \frac{ dy }{ dx }\]

Ok so this is where you got to I see and had trouble.

Answer 19

Yep, that's exactly what I did.

Answer 20

\[\huge -4x^3+8y^3\frac{ dy }{ dx }=3y+3x \frac{ dy }{ dx }\]

\[\huge -4x^3-3x \frac{ dy }{ dx }+8y^3\frac{ dy }{ dx }=3y\]

Answer 21

The whole point of these questions is to isolate dy/dx, can you finish it off?

Answer 22

I still get what I was getting before:
\[8y^3\frac{ dy }{ dx } - 3x \frac{ dy }{ dx } = 3y+4x^3\]
\[(8y^3-3x)\frac{ dy }{ dx } = 3y + 4x^3\]
\[\frac{ dy }{ dx } = \frac{ 3y + 4x^3 }{ 8y^3-3x }\]

Answer 23

Check your signs

Answer 24

\[-4x^3-3y = 3x \frac{ dy }{ dx } -8y^3\frac{ dy }{ dx }\]

Answer 25

Then -4x^3 - 3y / 3x - 8y^3 = dy/dx.

Answer 26

Which is right. Why does it only work moving the dx/dy to the right first? Or does it not matter and the other answer would be correct anyway?

Answer 27

it's -3xdy/dx not -3y

Answer 28

Where?

Answer 29

You wrote -4x^3-3y

Answer 30

Oh hold on I see what you're doing

Answer 31

\[\huge -4x^3-3x \frac{ dy }{ dx }+8y^3\frac{ dy }{ dx }=3y\]

Answer 32

You're just mixing up the algebra

Answer 33

How would you do it?

Answer 34

\[\huge \huge -4x^3+8y^3\frac{ dy }{ dx }=3y+3x \frac{ dy }{ dx }\]

Ok you got to this point, agree?

Answer 35

yep.

Answer 36

Ok so isolate dy/dx by moving it to the left side.

Answer 37

And the -4x^3 to the right?

Answer 38

Yes, but with these problems, take it step by step, it's very easy to make mistakes.

Answer 39

Okay :)

Answer 40

Okay :)\[-4x^3 + 8y^3\frac{ dy }{ dx } - 3x \frac{ dy }{ dx } = 3y\]

Answer 41

Good, continue :)

Answer 42

\[8y^3\frac{ dy }{ dx } - 3x \frac{ dy }{ dx } = 3y + 4x^3\]

Answer 43

Then isolating dy/dx:

Answer 44

Correct

Answer 45

\[(8y^3-3x)\frac{ dy }{ dx } = 3y + 4x^3\]

Answer 46

Good! Finish it off!

Answer 47

\[\frac{ dy }{ dx } = \frac{ 3y + 4x^3 }{ 8y^3 - 3x }\]

Answer 48

Good job!

Answer 49

Thank you!

Answer 50

Your welcome :)!