#### To determine

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

#### Answer

The implicit differentiation of x+y=1 is dydx=−yx.

#### Explanation

**Given:**

The equation x+y=1

**Derivative rules:**

(1) Chain rule: if y=f(u) and u=g(x) are both differentiable function, then

dydx=dydu⋅dudx

(2) Power rule: ddx(xn)=nxn−1

(3) Difference rule: ddx(f−g)=ddx(f)−ddx(g)

(4) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(5) Constant multiple rule: ddx(c⋅f)=c⋅ddx(f)

**Calculation:**

Obtain the derivative of the given equation.

x+y=1

Differentiate with respect to *x* on both sides,

ddx(x+y)=ddx(1)

Apply the derivative rules (4),(1) and (2),

ddx(x)+ddx(y)=ddx(1)ddx(x12)+ddx(y12)=ddx(1)(12x12−1)+(12y12−1)dydx=0(12x−12)+(12y−12)dydx=0

Simplify further and obtain the value dydx,

dydx=−12x−1212y−12=−y12x12=−yx

Therefore, the differentiation of x+y=1 is dydx=−yx.

#### To determine

**To find:** The equation explicitly for *y* and dydx.

#### Answer

The derivative of the equation x+y=1 is dydx=−1x+1.

#### Explanation

**Given:**

The equation x+y=1.

**Calculation:**

The given equation can be expressed explicitly for *y* as follows,

y=1−x

Take square on both sides,

y=(1−x)2=1+(x)2−2x=1+x−2x12

Apply the derivative rules (3),(4) and (5),

dydx=ddx(1+x−2x12)=ddx(1)+ddx(x)−ddx(2x12)=ddx(1)+ddx(x)−2ddx(x12)

Apply the derivative rule (2) and simplify further,

dydx=0+1−2(12x12−1)=1−2(12x−12)=−1x+1

Therefore, the derivative of the equation x+y=1 is dydx=−1x+1.

#### To determine

**To check:** Whether the solutions from part (a) and part (b) are consistent or not.

#### Answer

The solutions from part (a) and part (b) are consistent.

#### Explanation

The derivative of the equation y=1−x from part (a),

dydx=−yx

Substitute the value y=1−x (from part (b)),

dydx=−(1−x)x=−1x+xx=−1x+1

Since, the derivative of the equation x+y=1 from part (a) is same as the derivative of the equation x+y=1 from part (b).

Therefore, it can be concluded that the two solutions from part (a) and (b) are consistent.