Question

How do you find the derivative of `y^2-3xy+x^2=7`?

Answer 1

Differentiate each term.
When a term is a function of y, treat `dy/dx` as a dependent variable.
After differentiating, solve for `dy/dx`.

Explanation:

Given:

`y^2-3xy+x^2=7`

Differentiate each term:

`(d(y^2))/dx-(d(3xy))/dx+(d(x^2))/dx=(d(7))/dx`

The derivative of the constant term is 0:

`(d(y^2))/dx-(d(3xy))/dx+(d(x^2))/dx=0`

For the first term, I shall use the chain rule, `(d(f(y)))/dx = (d(f(y)))/dy dy/dx`:

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

Returning to the equation:

`2ydy/dx-(d(3xy))/dx+(d(x^2))/dx=0`

For the second term, I shall use the linear property of the derivative and the product rule:

`-(d(3xy))/dx = -3(d(xy))/dx = -3((d(x))/dxy+ xdy/dx) = -3y-3xdy/dx`

Returning to the equation:

`2ydy/dx-3y-3xdy/dx+(d(x^2))/dx=0`

For the third term, I shall use the power rule, `(d(x^n))/dx = nx^(n-1)`:

`2ydy/dx-3y-3xdy/dx+2x =0`

Solve for `dy/dx`:

`2ydy/dx-3xdy/dx =3y-2x`

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

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