$$\text { Use implicit differentiation to find } d y / d x \text { and } \mathrm { d } ^ { 2 } y / d x ^ { 2 } \text {. }$$
-$x y - x + y = 4$
A) $\frac { d y } { d x } = \frac { 1 - y } { 1 + x } ; \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 2 y - 2 } { ( x + 1 ) ^ { 2 } }$
B) $\frac { d y } { d x } = - \frac { 1 + y } { x + 1 } ; \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 2 y - 2 } { ( x + 1 ) ^ { 2 } }$
C) $\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { \mathrm { y } + 1 } { \mathrm { x } + 1 } ; \frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } \mathrm { x } ^ { 2 } } = \frac { 2 \mathrm { y } + 2 } { ( \mathrm { x } + 1 ) ^ { 2 } }$
D) $\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 1 + \mathrm { y } } { \mathrm { x } + 1 } ; \frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } 2 } = \frac { \mathrm { y } + 1 } { ( \mathrm { x } + 1 ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

We start by taking the implicit derivative of the given equation with respect to $x$: $$\frac{d}{dx}(xy-x+y)=\frac{d}{dx}4$$ Using the product rule and chain rule, we get: $$y+x\frac{dy}{dx}-1+\frac{dy}{dx}=0$$ Simplifying the equation, we get: $$\frac{dy}{dx}=\frac{1-y}{1+x}$$ Now, taking the derivative of $\frac{dy}{dx}$ with respect to $x$ using quotient rule, we get: $$\frac{d}{dx}\left(\frac{1-y}{1+x}\right)=\frac{\frac{d}{dx}(1-y)(1+x)-\frac{d}{dx}(1+x)(1-y)}{(1+x)^2}$$ Simplifying the above equation and using the previously derived equation for $\frac{dy}{dx}$, we get: $$\frac{d^2y}{dx^2}=\frac{2y-2}{(x+1)^2}$$ Therefore, the answer is A.

Use implicit differentiation to find dy/dx and d2y/dx2. \text { Use implicit differentiation to find } d y / d x \text { and } \mathrm { d } ^ { 2 } y / d x ^ { 2 } \text {. } Use implicit differentiation to find dy/dx and d2y/dx2.

$\text { Use implicit differentiation to find } d y / d x \text { and } \mathrm { d } ^ { 2 } y / d x ^ { 2 } \text {. }$