How do you implicitly differentiate #xy^2- yln(x-y)^2= 4-x#?

Question

Elijah Abram · Accepted Answer

Remember to use the chain rule and product rule when differentiating the y variable ...

#xy^2- yln(x-y)^2= 4-x# Now, differentiate ... #y^2+x(2y)y'-ln(x-y)^2-y/(x-y)^2(2)(x-y)(1-y')=-1# Next simplify and solve for #y'# ... #y'[(2xy)+(2y)/(x-y)]=ln(x-y)^2+(2y)/(x-y)-1# #y'=[ln(x-y)^2+(2y)/(x-y)-1]/[(2xy)+(2y)/(x-y)]# hope that helped

Ezra Adams · Answer

undefined To implicitly differentiate the given equation $xy^2 - y\ln(x-y)^2 = 4 - x$, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the product rule for the term $xy^2$ and the chain rule for the term $y\ln(x-y)^2$.
3. Simplify the resulting expression.

The implicit differentiation of the equation $xy^2 - y\ln(x-y)^2 = 4 - x$ with respect to $x$ yields:

$$y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - y\frac{2(x-y)}{x-y}\frac{dy}{dx} = -1$$

Simplify the expression:

$$y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - 2y\frac{x-y}{x-y}\frac{dy}{dx} = -1$$

$$y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - 2y\frac{dy}{dx} = -1$$

$$y^2 - \ln(x-y)^2 = 4 - x$$

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

$$y^2 - \ln(x-y)^2 = 4 - x$$