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

Question

Isabella Ackerman · Accepted Answer

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

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)#

Christopher Allers · Answer

undefined To find the derivative of $ y^2 - 3xy + x^2 = 7 $, differentiate each term with respect to $ x $ using the chain rule and product rule where necessary. This yields:

$$ \frac{d}{dx}(y^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(x^2) = \frac{d}{dx}(7) $$

$$ 2y\frac{dy}{dx} - (3x\frac{dy}{dx} + 3y) + 2x = 0 $$

$$ 2y\frac{dy}{dx} - 3x\frac{dy}{dx} - 3y + 2x = 0 $$

$$ (2y - 3x)\frac{dy}{dx} = 3y - 2x $$

$$ \frac{dy}{dx} = \frac{3y - 2x}{2y - 3x} $$