Question

find dy/dx if 2y-xy^3=8

Answer 1

Do implicit differentiation first.

Answer 2

is this first step correct?
2dy/dx-(dy/dx y^3+x3xy^2 dy/dx)=0

Answer 3

?? Looks like you're a little happy with the dy/dx. It's all about the Chain Rule!

\(\dfrac{d}{dx}y = \dfrac{dy}{dx}\) - You got this piece.

\(\dfrac{d}{dx}(-xy^{3}) = (-x)\dfrac{d}{dx}y^{3} + y^{3}\dfrac{d}{dx}(-x) = (-x)(3y^{2})\dfrac{dy}{dx} + y^{3}(-1)\)

Answer 4

thank you!

Answer 5

After you do the implicit differentiation on the left equate it to 0 as the right hand side is a constant which when differentiated will yield zero. Then gather up all the dy/dx terms on the left and express it as dy/dx = some expression.

I prefer to use the notation y' to dy/dx in implicit differentiation as it is less of a clutter.

Answer 6

I understand the second part, but I'm confused on the implicit differentiation (I'm making corrections on a test I didn't very poorly on). Can you please show me how?

Answer 7

*did very poorly on, sorry

Answer 8

2y - xy^3 = 8

The first term is 2y. Differentiate it with respect to y. It will be 2dy/dx or 2y'

The next term is xy^3 (we will take care of the negative sign later on)
Differentiate xy^3. Apply product rule.
First term * derivative of second + second term * derivative of first.
x * (3(y^2)y') + (y^3)(1) = 3x(y^2)y' + y^3

Put them together: 2y' - (3x(y^2)y' + y^3) = 0
Gather y' terms together.
y'(2 - 3xy^2) - y^3 = 0
y'(2 - 3xy^2) = y^3
y' = y^3 / (2 - 3xy^2)

Answer 9

The second line should be: Differentiate it with respect to x (not y)

Answer 10

In the following examples we are differentiating with respect to x:
Derivative of x is 1
Derivative of x^2 is 2x
Derivative of x^7 is 7x^6

Derivative of y is dy/dx (or y')
Derivative of y^2 is (2y)dy/dx = 2yy'
Derivative of y^7 is 7(y^6)dy/dx = 7(y^6)y'

Answer 11

Note: Derivative of y^7 with respect to y is 7y^6
But Derivative of y^7 with respect to x is 7y^6 * (dy/dx) (chain rule)

Answer 12

Thank you very much!! This was very helpful!

Answer 13

you are welcome. Glad to be able to assist.

Answer 14

\[x^n = nx^{n-1}\]