Implicit Differentiation Consider the equation: x

(1)

Consider the equation:

x2+y2=25 This equation describes a circle:

x y

0

This is not a function and we cannot write it as:

y = . . . unless we split the circle in upper and lower half

(2)

We can useimplicit differentiation:

I differentiate both sides of the equation w.r.t. x , and

I then solve for y0, that is, for dy_dx

We differentiate x2+y2=25 implicitly. We have d dx(x 2₊_y2_{) =} d dx25 d dxx 2₊ d dxy 2₌₀ 2x + d dxy

2₌₀ _{y is a function of x =}_{⇒ chain rule}

2x + d dy(y 2₎ d dxy = 0 2x + 2y d dxy = 0 =⇒ d dxy = − x y =⇒ dy dx = − x y

(3)

We can useimplicit differentiation:

I differentiate both sides of the equation w.r.t. x , and

I then solve for y0, that is, for dy_dx

We differentiate x2+y2=25 implicitly. We have dy

dx = − x y

Find an equation of the tangent at point (3, 4). At point (3, 4) we have:

dy dx = −

3 4 Thus the tangent is

y − 4 = −3

(4)

Note that the derivative now depends on x and y ! dy dx = − x y x y 0 slope −3₄ slope 3₄

(5)

Find y0 where x3+y3=6xy .

x y

(6)

Find y0 where x3+y3=6xy . d dx(x 3₊_y3_{) =} d dx6xy 3x2+3y2· y0 = d dx6xy 3x2+3y2· y0 =6x d dxy + y d dx6x

3x2+3y2· y0 =6xy0+6y we solve for y0

3y2· y0−6xy0 = +6y − 3x2 y0(3y2−6x ) = 6y − 3x2 y0 = 6y − 3x 2 3y2₋_6x = 2y − x2 y2₋_2x

(7)

y0 = 2y − x

2

y2₋_2x

Find the tangent to the curve at point (3, 3): y0= 2 · 3 − 3

2

32₋_{2 · 3} = −1

(8)

y0 = 2y − x

2

y2₋_2x

At what point in the first quadrant is the tangent horizontal? In the first quadrant x > 0 and y > 0, and

2y − x2=0 =⇒ y = x 2 2 =⇒ x 3_{+ (}x2 2 ) 3₌_6xx2 2 =⇒ x 6 8 −2x 3₌₀ ₌_⇒ _x3₍x3 8 −2) = 0 Since x > 0, we get x =√316. Then the point is

(9)

Find y0 where sin(x + y ) = y2cos x We have: d dx sin(x + y ) = d dx(y 2_{cos x )} cos(x + y ) · (1 + y0) =cos x · d dx(y 2_{) +}_y2 d dx(cos x ) cos(x + y ) + y0cos(x + y ) = cos x · (2yy0) +y2(−sin x )

y0cos(x + y ) − 2yy0cos x = −y2sin x − cos(x + y ) y0(cos(x + y ) − 2y cos x ) = −(y2sin x + cos(x + y ))

y0 = −y

2_{sin x + cos(x + y )}

(10)

Find y00where x4+y4=16 We have: d dx(x 4₊_y4_{) =} d dx16 =⇒ 4x 3₊_4y3_y0 ₌₀ ₌_{⇒ y}0_{= −}x3 y3 Thus y00= d dx −x 3 y3 = −y 3 d dxx 3₋_x3 d dxy 3 (y3₎2 = − y3_3x2₋_x3_3y2_y0 y6 = − 3x2y3−3x3y2−x3 y3 y6 = − 3x2(x4+y4) y7 = − 48x2 y7