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

1 Answer
Dec 14, 2017

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)#