Question

Question #ae837

Answer 1

Differentiate each term with respect to x:

`(d(x^3))/dx-(d(xy^2))/dx+(d(y^3))/dx=(d(1))/dx`

The first term requires the use of the power rule `(d(x^n))/dx = nx^(n-1)`:

`3x^2-(d(xy^2))/dx+(d(y^3))/dx=(d(1))/dx`

The second term requires the use of the product rule, `(d(uv))/dx = (du)/dxv+u(dv)/dx`:

`3x^2-(dx/dxy^2+x(d(y^2))/dx)+(d(y^3))/dx=(d(1))/dx`

`dx/dx = 1` and we use the chain rule `(d(u(v(x))))/dx = (du)/(dv)(dv)/dx` second term within the parenthesis:

`3x^2-(y^2+x(d(y^2))/dy dy/dx)+(d(y^3))/dx=(d(1))/dx`

Now, we use the power rule `(d(y^2))/dy = 2y`:

`3x^2-(y^2+2xy dy/dx)+(d(y^3))/dx=(d(1))/dx`

Distribute the minus sign:

`3x^2-y^2-2xy dy/dx+(d(y^3))/dx=(d(1))/dx`

Use the chain rule, `(d(u(v(x))))/dx = (du)/(dv)(dv)/dx`, for the 4th term:

`3x^2-y^2-2xy dy/dx+(d(y^3))/dy dy/dx=(d(1))/dx`

Use the power rule on the 4th term:

`3x^2-y^2-2xy dy/dx+3y^2 dy/dx=(d(1))/dx`

The derivative of a constant is 0:

`3x^2-y^2-2xy dy/dx+3y^2 dy/dx=0`

Move all of the terms that do not contain `dy/dx` to the right:

`-2xy dy/dx+3y^2 dy/dx=y^2-3x^2`

Factor `dy/dx` out of the terms on the left:

`(3y^2-2xy) dy/dx=y^2-3x^2`

Divide both sides by the leading factor:

`dy/dx=(y^2-3x^2)/(3y^2-2xy)`

Done.

Answer 2

`dy/dx=(y^2-3x^2)/(3y^2-2xy)`

Explanation:

`"differentiate "color(blue)"implicitly with respect to x"`

`"noting that "d/dx(y)=dy/dx" and"`

`d/dx(y^2)=2ydy/dx`

`"differentiate "xy^2" using the "color(blue)"product rule"`

`"given "y=f(x)g(x)" then"`

`dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"`

`rArr3x^2-(x.2ydy/dx+y^2)+3y^2dy/dx=0`

`rArr3x^2-2xydy/dx-y^2+3y^2dy/dx=0`

`rArrdy/dx(3y^2-2xy)=y^2-3x^2`

`rArrdy/dx=(y^2-3x^2)/(3y^2-2xy)`