#### To determine

**To find:** The derivative dydx by implicit differentiation.

#### Answer

The differentiation of x2+xy+y2+1=0 is. dydx=−2x−yx+2y.

#### Explanation

**Given:**

The equation x2+xy+y2+1=0.

**Derivative rules:**

(1) Chain rule: If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

(2) Product rule: ddx(fg)=fddx(g)+gddx(f)

(3). Power Rule: ddx(xn)=nxn−1

(4) Sum Rule: ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]

**Calculation:**

Obtain the derivative of x2+xy+y2+1=0 implicitly with respect to *x*.

Use Sum rule, product rule and power rule and differentiate with respect to *x* on both sides of the equation x2+xy+y2+1=0,

ddx(x2+xy+y2+1)=ddx(0)ddx(x2)+ddx(xy)+ddx(y2)+ddx(1)=02x2−1+[xddx(y)+yddx(x)]+2y2−1dydx=02x+xdydx+y(1)+2ydydx=0

Simplify further and obtain the derivative dydx as,

2x+xdydx+y+2ydydx=0xdydx+2ydydx=−2x−ydydx(x+2y)=−2x−ydydx=−2x−yx+2y

Therefore, the differentiation of x2+xy+y2+1=0 is. dydx=−2x−yx+2y.

#### To determine

**To sketch:** The curve in part (a) and prove that the observation obtained is correct.

#### Explanation

**Given:**

The equation x2+xy+y2+1=0.

**Calculation:**

Use online graphing calculator and sketch the curve x2+xy+y2+1=0 as shown below in Figure 1.

From the above Figure 1,it can be concluded that there is no point (x,y) which satisfies the equation x2+xy+y2+1=0.

To prove the observation, the equation x2+xy+y2+1=0 can be simplified as (x2+2xy2)+y2+1=0.

Add and subtract (y2)2 in the above expression as follows.

(x2+2xy2+(y2)2−(y2)2)+y2+1=0[x2+2xy2+(y2)2]−(y2)2+y2+1=0[x+y2]2−y24+y2+1=0[x+y2]2+3y24+1=0

Notice that the first term is a square term and is always positive, likewise the second term is also a square term and hence cannot be negative and therefore the left hand side will never become zero.

Therefore, there is no point (x,y) which satisfies the equation x2+xy+y2+1=0.

#### To determine

**To explain:** The observation of dydx in part (a) with the view of part (b).

#### Explanation

From part (b), it is clear that the equation x2+xy+y2+1=0 has no solution as there is no point (x,y) that satisfies the equation.

Hence in part (a), only the implicit differentiation was found out and it doesn’t holds any relevance to the solution of the equation x2+xy+y2+1=0.