Derivative of implicit equations

dy To find dx

dy To find dx From an implicit equation, differentiate each term

dy To find dx From an implicit equation, differentiate each term respect to x, using the chain rule

dy To find dx From an implicit equation, differentiate each term respect to x, using the chain rule to differentiate any function

dy To find dx From an implicit equation, differentiate each term respect to x, using the chain rule to differentiate any function

f (y)as

dy To find dx From an implicit equation, differentiate each term respect to x, using the chain rule to differentiate any function

f (y)as

Show that (1,2)

Show that (1,2) is on the circle

Show that (1,2) is on the circle x2 + y 2 − 6x + 2y − 3 = 0

Show that (1,2) is on the circle x2 + y 2 − 6x + 2y − 3 = 0

Show that (1,2) is on the circle x2 + y 2 − 6x + 2y − 3 = 0

dy and find dx at that point

Show that (1,2) is on the circle x2 + y 2 − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

Show that (1,2) is on the circle x2 + y 2 − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

12 + 22 − 6(1) + 2(2) − 3 = 0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

12 + 22 − 6(1) + 2(2) − 3 = 0 1+4−6+4−3=0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

12 + 22 − 6(1) + 2(2) − 3 = 0 1+4−6+4−3=0 5−6+4−3=0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

1+4−6+4−3=0 5−6+4−3=0 −1 + 4 − 3 = 0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

5−6+4−3=0 −1 + 4 − 3 = 0 3−3=0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

3−3=0 0=0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

3−3=0 0=0

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

3−3=0 0=0 True

Show that (1,2) is on the circle 2

2

x + y − 6x + 2y − 3 = 0 Step 1: Substitute the coordinates of the point

(1, 2)

↙

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0

term by term

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dx

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dx −6

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy +2 −6 dx

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx dx

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

Factor 2 out

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

dy dy 2(x + y dx − 3 + dx ) = 0

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

dy dy 2(x + y dx − 3 + dx ) = 0

2

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

dy dy 2(x + y dx − 3 + dx ) = 0

2

2

Differentiate ↓ x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

dy dy 2(x + y dx − 3 + dx ) = 0 dy dy x + y dx − 3 + dx = 0

Find the rate of change at (1,2) x2 + y 2 − 6x + 2y − 3 = 0 2x +2y

dy dy +2 −6 dx = 0 dx

dy dy x + y dx − 3 + dx = 0

Find the rate of change at (1,2) dy dy x + y dx − 3 + dx = 0

Find the rate of change at (1,2) dy dy x + y dx − 3 + dx = 0

Find the rate of change at (1,2) dy dy 1 + 2 dx − 3 + dx = 0

Find the rate of change at (1,2) dy dy 1 + 2 dx − 3 + dx = 0

3dy −2 + dx = 0

Find the rate of change at (1,2) 3dy −2 + dx = 0 3dy = 0 + 2 dx

Find the rate of change at (1,2) 3dy = 0 + 2 dx

3dy = 2 dx

Find the rate of change at (1,2) 3dy = 0 + 2 dx

3dy = 2 dx

dy 2 = dx 3