Question

dy/dx [5x^3+x^2y-xy^3=3]

Answer 1

@Easyaspi314 could you explain how to do this one. its more confusing

Answer 2

You need implicit differentiation. Keep in mind that you would need the Product Rule when finding the derivative of x^2y and xy^3.

We have 5x^3 + x^2y - xy^3 = 3

We will take the derivative of both sides of the equation.

15x^2 + (x^2)(y ') + (y)(2x) - (3xy^2 y' + y^3) = 0
15x^2 + x^2y' + 2xy - 3xy^2y' - y^3 = 0
x^2y' - 3xy^2y' = y^3 - 15x^2
factor out y', we get y'(x^2 - 3xy^2) = y^3 - 15x^2

Divide both sides by x^2 - 3xy^2, we get

y' = (y^3 - 15x^2)/ (x^2 - 3xy^2)

Answer 3

Makes sense? Any questions?

Answer 4

nope i think i get it. thanks

Answer 5

I used y' instead of dy/dx...they mean the same thing..many textbooks use dy/dx and many use y'..no difference whatsoever.

Answer 6

it says its wrong

Answer 7

whats wrong?

Answer 8

shouldn't this be (3xy^2 y' + y^3) be (x3y^2 y'+y^3) instead

Answer 9

Let me look at it........

Answer 10

No... I do not see any error.

Answer 11

its online hw so it tells me if i am right or wrong

Answer 12

I used the product rule to find the derivative of xy^3 and I put that derivative in parenthesis by itself.

Answer 13

so what would the answer be @Easyaspi314

Answer 14

answer will be what I have in my last step.

Final answer: y' = (y^3 - 15x^2)/ (x^2 - 3xy^2)

Answer 15

@Easyaspi314 is right

Answer 16

If I made an arithmetic error, please point it out to me. I dont think there's any error.

Answer 17

thats what i tried and it didnt work. i wouldnt know what is wrong

Answer 18

Again, I dont see anything wrong. Why do you think something is wrong?

Answer 19

i dont know. its ok i will ask my professor how to do it tomorrow

Answer 20

why is the y^3 positive instead of negative. at the second like you turn it negative, but at the first line it's positive @Easyaspi314

Answer 21

I put a minus sign outside of the parenthesis, so it really is negative.

Answer 22

In other words when you see -xy^3 in the equation ...I put a minus sign, then I opened up parenthesis and took the derivative of just xy^3. So what is inside the parenthesis is just the derivative of xy^3. Look at it again. I did it that way so as to not be confused when taking the derivative of -xy^3.