implicit (dy)/(dx),x^6-5xy^3=9xy

⚠ xy is read as the product x*y: for the variable xy, write xy_

1) For

x^{6} - 5 x y^{3} = 9 x y

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

Set F(x, y) =

x^{6} - 5 x y^{3} - 9 x y

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

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

⇥1.1) Let's perform the differentiation:

\frac{d}{d x} (x^{6} - 5 x y^{3} - 9 x y)

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

\frac{d}{d x} (x^{6} - 5 x y^{3} - 9 x y)

\frac{d}{d x} (x^{6}) + \frac{d}{d x} (- 5 x y^{3}) + \frac{d}{d x} (- 9 x y)

⇥

\frac{d}{d x} (x^{6}) + \frac{d}{d x} (- 5 x y^{3}) + \frac{d}{d x} (- 9 x y)

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

\frac{d}{d x} (x^{6})

6 x^{5}

■For differentiation with respect to

x

-5 y^{3}

is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (- 5 x y^{3})

-5 y^{3} \cdot \frac{d}{d x} (x)

■For differentiation with respect to

x

-9 y

is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (- 9 x y)

-9 y \cdot \frac{d}{d x} (x)

⇥

6 x^{5} - 5 y^{3} \cdot \frac{d}{d x} (x) - 9 y \cdot \frac{d}{d x} (x)

⇥ⓘ The derivative of x is 1:

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

1

⇥

6 x^{5} - 5 y^{3} \cdot 1 - 9 y \cdot 1

⇥

6 x^{5} - 5 y^{3} - 9 y

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

⇥1.2) Let's differentiate:

\frac{d}{d y} (x^{6} - 5 x y^{3} - 9 x y)

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

\frac{d}{d y} (x^{6} - 5 x y^{3} - 9 x y)

\frac{d}{d y} (x^{6}) + \frac{d}{d y} (- 5 x y^{3}) + \frac{d}{d y} (- 9 x y)

⇥

\frac{d}{d y} (x^{6}) + \frac{d}{d y} (- 5 x y^{3}) + \frac{d}{d y} (- 9 x y)

⇥ⓘ The derivative of a constant is zero:

\frac{d}{d y} (x^{6})

0

■For differentiation with respect to

y

-5 x

is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d y} (- 5 x y^{3})

-5 x \cdot \frac{d}{d y} (y^{3})

■For differentiation with respect to

y

-9 x

is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d y} (- 9 x y)

-9 x \cdot \frac{d}{d y} (y)

⇥

0 - 5 x \cdot \frac{d}{d y} (y^{3}) - 9 x \cdot \frac{d}{d y} (y)

⇥ⓘ Adding 0 changes nothing:

0 - 5 x \cdot \frac{d}{d y} (y^{3})

-5 x \cdot \frac{d}{d y} (y^{3})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d y} (y^{3})

3 y^{2}

■The derivative of x is 1:

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

1

⇥

- 5 x \cdot 3 y^{2} - 9 x \cdot 1

⇥ⓘ Multiplications are done before additions and subtractions:

5 \cdot 3

15

⇥

- 15 x y^{2} - 9 x

The partial derivatives are Fx =

6 x^{5} - 5 y^{3} - 9 y

and Fy =

- 15 x y^{2} - 9 x

- 15 x y^{2} - 9 x \neq 0

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

⇥

- \frac{6 x^{5} - 5 y^{3} - 9 y}{- 15 x y^{2} - 9 x}

⇥ⓘ The minus sign is factored out, changing the sign of each term:

- 15 x y^{2} - 9 x

- (15 x y^{2} + 9 x)

⇥

- \frac{6 x^{5} - 5 y^{3} - 9 y}{- (15 x y^{2} + 9 x)}

⇥ⓘ Two minus signs cancel out:

- \frac{6 x^{5} - 5 y^{3} - 9 y}{- (15 x y^{2} + 9 x)}

\frac{6 x^{5} - 5 y^{3} - 9 y}{15 x y^{2} + 9 x}

⇥

\frac{6 x^{5} - 5 y^{3} - 9 y}{15 x y^{2} + 9 x}

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{6 x^{5} - 5 y^{3} - 9 y}{15 x y^{2} + 9 x}

\frac{6 x^{5} - 5 y^{3} - 9 y}{3 x \cdot (5 y^{2} + 3)}

▶Implicit derivative dy/dx:

\frac{6 x^{5} - 5 y^{3} - 9 y}{3 x \cdot (5 y^{2} + 3)}