Chain Rule/Directional derivatives - Department of ...ccroke/lecture14-4,5.pdfChain Rule/Directional derivatives Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011

Chain Rule/Directional derivatives

Christopher Croke

University of Pennsylvania

Math 115UPenn, Fall 2011

Christopher Croke Calculus 115

Chain Rule

In the one variable case z = f (y) and y = g(x) then dzdx = dz

dydydx .

When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t. and

dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt.

or in other words

dw

dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).

The proof is not hard and given in the text.Show tree diagram.


Chain Rule


dydydx .

When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t.

and

dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt.

or in other words

dw




Chain Rule


dydydx .


dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt.

or in other words

dw




Chain Rule


dydydx .


dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt.

or in other words

dw




Chain Rule


dydydx .


dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt.

or in other words

dw


The proof is not hard and given in the text.

Show tree diagram.


Chain Rule


dydydx .


dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt.

or in other words

dw




Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),

and y = sin(t).

What is dwdt (π2 )?

Do tree diagram for w=f(x,y,z) where x ,y ,z are functions of t.

What do you do if the intermediate variables are also functions oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ. Forexample compute ∂w

∂r .

Problem: Compute ∂w∂r and ∂w

∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .



and y = sin(t).What is dw

dt (π2 )?



∂r .






dt (π2 )?



∂r .






dt (π2 )?


What do you do if the intermediate variables are also functions oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ.

Forexample compute ∂w

∂r .






dt (π2 )?



∂r .






dt (π2 )?



∂r .




Implicit Differentiation

If F (x , y) = 0 defines y implicitly as a function of x , that isy = h(x), and if F is differentiable then:

0 =dF

dx= Fx · 1 + Fy ·

dy

dx.

Thusdy

dx= −Fx

Fy

whenever Fy 6= 0

Problem: Find dydx if x2 + y2x + sin(y) = 0.




0 =dF

dx= Fx · 1 + Fy ·

dy

dx.

Thusdy

dx= −Fx

Fy

whenever Fy 6= 0





0 =dF

dx= Fx · 1 + Fy ·

dy

dx.

Thusdy

dx= −Fx

Fy

whenever Fy 6= 0



Directional Derivatives

Given a function f (x , y) then the rate of change (w.r.t.t) along acurve (x(t), y(t)) is (by the chain rule)

fxdx

dt+ fy

dy

dt.

If ~u is a unit vector we can write ~u = u1~i + u2~j (whereu21 + u22 = 1). So the line through (x0, y0) in the direction ~u witharc length parameter is (x(s), y(s)) where:

x(s) = u1s + x0 y(s) = u2s + y0.

The directional derivative in the direction ~u is thus(dfds

)~u,(x0,y0)

= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf

)(x0,y0)

.




fxdx

dt+ fy

dy

dt.

If ~u is a unit vector we can write ~u = u1~i + u2~j (whereu21 + u22 = 1).

So the line through (x0, y0) in the direction ~u witharc length parameter is (x(s), y(s)) where:

x(s) = u1s + x0 y(s) = u2s + y0.


)~u,(x0,y0)

= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf

)(x0,y0)

.




fxdx

dt+ fy

dy

dt.


x(s) = u1s + x0 y(s) = u2s + y0.


)~u,(x0,y0)

= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf

)(x0,y0)

.




fxdx

dt+ fy

dy

dt.


x(s) = u1s + x0 y(s) = u2s + y0.


)~u,(x0,y0)

= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf

)(x0,y0)

.


Definition:(D~uf

)(x0,y0)

= lims→0

f (x0 + u1s, y0 + u2s)− f (x0, y0)

s.

What is geometric meaning?

If you look at the formula you note that(D~uf

)(x0,y0)

=(fx(x0, y0)~i + fy (x0, y0)~j

)· ~u

We call the vector fx(x0, y0)~i + fy (x0, y0)~j the gradient of f at(x0, y0) and we denote it ∇f .So

∇f = fx~i + fy~j .

andD~uf = ∇f · ~u.


Definition:(D~uf

)(x0,y0)

= lims→0

f (x0 + u1s, y0 + u2s)− f (x0, y0)

s.



)(x0,y0)

=(fx(x0, y0)~i + fy (x0, y0)~j

)· ~u


∇f = fx~i + fy~j .



Definition:(D~uf

)(x0,y0)

= lims→0

f (x0 + u1s, y0 + u2s)− f (x0, y0)

s.



)(x0,y0)

=(fx(x0, y0)~i + fy (x0, y0)~j

)· ~u


∇f = fx~i + fy~j .



Definition:(D~uf

)(x0,y0)

= lims→0

f (x0 + u1s, y0 + u2s)− f (x0, y0)

s.



)(x0,y0)

=(fx(x0, y0)~i + fy (x0, y0)~j

)· ~u

We call the vector fx(x0, y0)~i + fy (x0, y0)~j the gradient of f at(x0, y0) and we denote it ∇f .

So∇f = fx~i + fy~j .



Definition:(D~uf

)(x0,y0)

= lims→0

f (x0 + u1s, y0 + u2s)− f (x0, y0)

s.



)(x0,y0)

=(fx(x0, y0)~i + fy (x0, y0)~j

)· ~u


∇f = fx~i + fy~j .



Definition:(D~uf

)(x0,y0)

= lims→0

f (x0 + u1s, y0 + u2s)− f (x0, y0)

s.



)(x0,y0)

=(fx(x0, y0)~i + fy (x0, y0)~j

)· ~u


∇f = fx~i + fy~j .



Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√

13~i − 3√

13~j .

If θ is the angle between ∇f and ~u then

D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).

This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).

Another conclusion is that if ~u⊥∇f then D~uf = 0



13~i − 3√

13~j .







13~i − 3√

13~j .



This means that f increases most rapidly in the direction of ∇f .

(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).




13~i − 3√

13~j .



This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)

In this direction D~uf = |∇f | (or −|∇f |).




13~i − 3√

13~j .







13~i − 3√

13~j .






Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?

Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:

d

dtf (x(t), y(t)) = fx ·

dx

dt+ fy ·

dy

dt= 0.

In other words:

∇f · (dxdt~i +

dy

dt~j) = 0.

Which says that ∇f is perpendicular to the level curve. So thetangent line to the level curve is the line perpendicular to ∇f .

Problem: Find tangent line to the curve

x + sin(y) + exy = 2

at the point (1, 0).




d

dtf (x(t), y(t)) = fx ·

dx

dt+ fy ·

dy

dt= 0.

In other words:

∇f · (dxdt~i +

dy

dt~j) = 0.








d

dtf (x(t), y(t)) = fx ·

dx

dt+ fy ·

dy

dt= 0.

In other words:

∇f · (dxdt~i +

dy

dt~j) = 0.








d

dtf (x(t), y(t)) = fx ·

dx

dt+ fy ·

dy

dt= 0.

In other words:

∇f · (dxdt~i +

dy

dt~j) = 0.

Which says that ∇f is perpendicular to the level curve.

So thetangent line to the level curve is the line perpendicular to ∇f .







d

dtf (x(t), y(t)) = fx ·

dx

dt+ fy ·

dy

dt= 0.

In other words:

∇f · (dxdt~i +

dy

dt~j) = 0.








d

dtf (x(t), y(t)) = fx ·

dx

dt+ fy ·

dy

dt= 0.

In other words:

∇f · (dxdt~i +

dy

dt~j) = 0.




at the point (1, 0).Christopher Croke Calculus 115

Three variables

Lets look at three independent variables, i.e. f (x , y , z). Then

∇f = fx~i + fy~j + fz~k .

andD~uf = ∇f · ~u = |∇f | cos(θ).

where θ is the angle between ∇f and ~u.

This means that ∇f is perpendicular to the level surfaces of f .(i..e it is normal to all curves in the level surface.)

The Tangent Plane at (x0, y0, z0) to the level surfacef (x , y , z) = c is the plane through the point (x0, y0, z0) normal to∇f (x0, y0, z0).Thus the equation is:

fx(x − x0) + fy (y − y0) + fz(z − z0) = 0.


Three variables



andD~uf = ∇f · ~u = |∇f | cos(θ).




fx(x − x0) + fy (y − y0) + fz(z − z0) = 0.


Three variables



andD~uf = ∇f · ~u = |∇f | cos(θ).




fx(x − x0) + fy (y − y0) + fz(z − z0) = 0.


Three variables



andD~uf = ∇f · ~u = |∇f | cos(θ).



The Tangent Plane at (x0, y0, z0) to the level surfacef (x , y , z) = c is the plane through the point (x0, y0, z0) normal to∇f (x0, y0, z0).

Thus the equation is:

fx(x − x0) + fy (y − y0) + fz(z − z0) = 0.


Three variables



andD~uf = ∇f · ~u = |∇f | cos(θ).




fx(x − x0) + fy (y − y0) + fz(z − z0) = 0.


The Normal Line of the surface at (x0, y0, z0) is the line through(x0, y0, z0) which is parallel to ∇f .

x = x0 + fx t, y = y0 + fy t, z = z0 + fz t.

Problem: Find the tangent plane and normal line to the graph ofz = f (x , y) = x2 − 3xy at the point (1, 2,−5).










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Chain Rule/Directional derivatives - Department of ...ccroke/lecture14-4,5.pdfChain Rule/Directional derivatives Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011