How do you implicitly differentiate # y^2+(y-x)^2-y/x^2-3y#?

Question

How do you implicitly differentiate #  y^2+(y-x)^2-y/x^2-3y#?

Luke Alvarez · Accepted Answer

First expand the brackets to give:
#y^2+(y-x)^2-y/x^2-3y#
#=y^2+y^2-2xy+x^2-y/x^2-3y#
#=2y^2-2xy+x^2-yxxx^-2-3y# By implicit differentiation (presumably with respect to x here):
#d/dx y^a=ay^(a-1)xxdy/dx#.
Also using the product rule:
If #f(x) = uv#, then #f'(x)=u'v+uv'# So the original expression differentiates to:
#4y^1 dy/dx - 2(xdy/dx+yxx1)+2x^1-(-2x^-3y+x^-2dy/dx)-3y^0dy/dx#
#=4ydy/dx-2xdy/dx-2y+2x+(2y)/x^3-1/x^2dy/dx#

Evelyn Agard · Answer

undefined To implicitly differentiate $ y^2+(y-x)^2-\frac{y}{x^2}-3y $ with respect to $ x $, you would apply the chain rule and product rule as needed. The steps are as follows:

1. Differentiate each term with respect to $ x $.
2. Use the chain rule for terms involving $ y $ to differentiate the outer function and then multiply by the derivative of the inner function.
3. Simplify the result.

The derivative of the given expression is:

$$ \frac{d}{dx} \left( y^2 + (y-x)^2 - \frac{y}{x^2} - 3y \right) = 2y\frac{dy}{dx} + 2(y-x)(\frac{dy}{dx}-1) - \frac{x^2y' - 2xy}{x^4} - 3 $$

$$ = 2y\frac{dy}{dx} + 2(y-x)\frac{dy}{dx} - 2(y-x) - \frac{x^2y' - 2xy}{x^4} - 3 $$

$$ = 2y\frac{dy}{dx} + 2y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y + 2x - \frac{x^2y' - 2xy}{x^4} - 3 $$

$$ = 4y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y - \frac{x^2y' - 2xy}{x^4} + 2x - 3 $$

$$ = 4y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y - \frac{y'}{x^2} + \frac{2y}{x^3} + 2x - 3 $$

$$ = (4y - 2x)\frac{dy}{dx} - 2y + \frac{2y}{x^3} + 2x - 3 - \frac{y'}{x^2} $$

So, the implicit derivative is $ (4y - 2x)\frac{dy}{dx} - 2y + \frac{2y}{x^3} + 2x - 3 - \frac{y'}{x^2} $.