What is the slope of the tangent line of #xy^2-(1-xy)^2= C #, where C is an arbitrary constant, at #(1,-1)#?

Question

Isaiah Allen · Accepted Answer

#dy/dx=-1.5# We first find #d/dx# of each term. #d/dx[xy^2]-d/dx[(1-xy)^2]=d/dx[C]# #d/dx[x]y^2+d/dx[y^2]x-2(1-xy)d/dx[1-xy]=0# #y^2+d/dx[y^2]x-2(1-xy)(d/dx[1]-d/dx[xy])=0# #y^2+d/dx[y^2]x-2(1-xy)(-d/dx[x]y+d/dx[y]x)=0# #y^2+d/dx[y^2]x-2(1-xy)(-y+d/dx[y]x)=0# The chain rule tells us: #d/dx=d/dy*dy/dx# #y^2+dy/dx d/dy[y^2]x-2(1-xy)(-y+dy/dxd/dy[y]x)=0# #y^2+dy/dx 2yx-2(1-xy)(-y+dy/dx x)=0# #dy/dx 2yx-2(1-x)dy/dx x=-y^2-2y(1-xy)# #dy/dx( 2yx-2x(1-x))=-y^2-2y(1-xy)#x #dy/dx=-(y^2+2y(1-xy))/(2yx-2x(1-x))# For #(1,-1)# #dy/dx=-((-1)^2+2(-1)(1-1(-1)))/(2(1)(-1)-2(1)(1-1))=-1.5#

William Abdalla · Answer

undefined To find the slope of the tangent line at the point (1, -1) of the curve $xy^2 - (1 - xy)^2 = C$, we first need to find the derivative of the curve with respect to $x$, then evaluate it at the point (1, -1).

Given the curve equation $xy^2 - (1 - xy)^2 = C$, take the derivative with respect to $x$:
$$\frac{d}{dx}[xy^2 - (1 - xy)^2] = \frac{d}{dx}[C]$$

Differentiate each term using the chain rule and product rule:

$$\frac{d(xy^2)}{dx} - \frac{d((1 - xy)^2)}{dx} = 0$$

Now, differentiate each term separately:

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

Simplify and collect terms:

$$y^2 + 2xy\frac{dy}{dx} - 2(1 - xy)(y) - 2x(y)\frac{dy}{dx} = 0$$

Now, plug in $x = 1$ and $y = -1$ into the equation and solve for $\frac{dy}{dx}$:

$$(-1)^2 + 2(1)(-1)\frac{dy}{dx} - 2(1 - (1)(-1))(-1) - 2(1)(-1)\frac{dy}{dx} = 0$$

Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{1 - 2}{2} = -\frac{1}{2}$$

Therefore, the slope of the tangent line at the point (1, -1) is $-\frac{1}{2}$.