implicit x^2+y^2=1 — step-by-step solution

1) For

x^{2} + y^{2} = 1

, treat y as a differentiable function of x and differentiate both sides

Set F(x, y) =

x^{2} + y^{2} - 1

. The chain rule gives Fx + Fy*dy/dx = 0

For Fx, differentiate with respect to x while holding y constant

⇥1.1) Let's differentiate:

\frac{d}{d x} (x^{2} + y^{2} - 1)

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x^{2} + y^{2} - 1)

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) + 0

⇥

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) + 0

⇥ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{2})

2 x

■Adding 0 changes nothing:

\frac{d}{d x} (y^{2}) + 0

\frac{d}{d x} (y^{2})

■The derivative of a constant is zero:

\frac{d}{d x} (y^{2})

0

⇥

2 x + 0

⇥ⓘ Adding 0 changes nothing:

2 x + 0

2 x

⇥

2 x

For Fy, differentiate with respect to y while holding x constant

⇥1.2) Let's perform the differentiation:

\frac{d}{d y} (x^{2} + y^{2} - 1)

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d y} (x^{2} + y^{2} - 1)

\frac{d}{d y} (x^{2}) + \frac{d}{d y} (y^{2}) + 0

⇥

\frac{d}{d y} (x^{2}) + \frac{d}{d y} (y^{2}) + 0

⇥ⓘ The derivative of a constant is zero:

\frac{d}{d y} (x^{2})

0

■Adding 0 changes nothing:

\frac{d}{d y} (y^{2}) + 0

\frac{d}{d y} (y^{2})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d y} (y^{2})

2 y

⇥

0 + 2 y

⇥ⓘ Adding 0 changes nothing:

0 + 2 y

2 y

⇥

2 y

The partial derivatives are Fx =

2 x

and Fy =

2 y

2 y \neq 0

, isolate the derivative: dy/dx = -Fx/Fy. This condition is taken on the original curve

⇥

- \frac{2 x}{2 y}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

- \frac{2 x}{2 y}

- \frac{x}{y}

▶Implicit derivative dy/dx:

- \frac{x}{y}