PARTIAL DERIVATIVES

MATH 200 WEEK 4 - MONDAY

MATH 200

GOALS

▸ Figure out how to take derivatives of functions of multiple variables and what those derivatives mean.

▸ Be able to compute first-order and second-order partial derivatives.

▸ Be able to perform implicit partial differentiation.

▸ Be able to solve various word problems involving rates of change, which use partial derivatives.

MATH 200

REVIEW▸ Calculus I:

▸ dy/dx = f’(x)

▸ f’(a) = “slope of the tangent line to f at x = a”

▸ e.g.f(x) = x2 − 1

f ′(x) = 2xf ′(1) = 2(1) = 2

f(1) = (1)2 − 1 = 0THE POINT (1,0) IS ON THE GRAPH OF f

THE SLOPE OF THE LINE TANGENT TO F AT (1,0) IS 2

MATH 200

CONSIDER A 3D EXAMPLE▸ Let f(x,y) = x2 + y2

▸ Consider the trace of f on the plane y=1

▸ f(x,1) = x2 + 1

MATH 200

▸ We can certainly find slope of the line tangent to z = x2+1 at any point on the xz-plane…

▸ Differentiating with respect to x, we get dz/dx = 2x

▸ The slope of the line at x=1 is 2

MATH 200

▸ So, can we write parametric equations or a vector-valued function for that same line in 3-Space…?

▸ Need: (1) a point on the line and (2) a direction vector parallel to the line

▸ We were at x=1 on the plane y=1. Since f(1,1) = 2, the point of tangency is (1,1,2)

▸ The slope of 2 is telling us [change in z]/[change in x]

▸ We need a direction vector for which z/x = 2 and y=0…

▸ <1,0,2> works! (There are infinitely many other choices of course.

L :

⎧⎪⎨

⎪⎩

x = 1 + t

y = 1

z = 2 + 2t

MATH 200

▸ What if we repeat the same process for the trace of f on the plane y=2?

▸ We’d be looking at the trace f(x,2) = x2 + 4

▸ Here it is on the xz-plane with its tangent line at x=1

MATH 200

▸ We can find slope of the line tangent to z = x2+4 at any point on the xz-plane…

▸ Differentiating with respect to x, we get dz/dx = 2x

▸ It’s the same!

▸ The slope of the line at x=1 is 2

▸ Since [change in z]/[change in x] = 2, we can use the direction vector <1,0,2> again, with starting point (1,2,5)

L :

⎧⎪⎨

⎪⎩

x = 1 + t

y = 2

z = 5 + 2t

MATH 200

WHAT JUST HAPPENED…?▸ We just our first partial derivative!

▸ Notice that in both cases (whether we set y=1 or y=2) we got dz/dx = 2x

▸ This would’ve been the case with any choice of constant value for y

▸ We could have done the same work on any plane of the form x=constant

▸ In that case, we’d find [change in z]/[change in y]

MATH 200

DEFINITIONS▸ The partial derivative of f with respect to x is what you get

when you…

▸ treat y as a constant…

▸ and differentiate with respect to x

▸ We write any of the following:

∂f

∂x,

∂z

∂x, fx(x, y)

WE USE THIS PARTIAL SYMBOL INSTEAD OF JUST d TO INDICATE

THAT THERE IS MORE THAN ONE INDEPENDENT

VARIABLE

∂

MATH 200

DEFINITIONS▸ The partial derivative of f with respect to y is what you get

when you…

▸ treat x as a constant…

▸ and differentiate with respect to y

▸ We write any of the following:

∂f

∂y,

∂z

∂y, fy(x, y)

NOTICE: WE’RE NOT USING “PRIME” NOTATION

ANYMORE…

IF I WRITE f’(x,y), YOU DON’T KNOW WHICH

VARIABLE I’M HOLDING CONSTANT

MATH 200

A LITTLE PRACTICE▸ Compute both first-order partial derivatives (fancy way of

saying first derivatives) for the following functions

▸ i.e. compute the partial derivative with respect to x and the partial derivative with respect to y for each function

1. f(x, y) = 3x2 � 2y + 1

2. g(x, y) = xy2 � ln y

3. h(x, y) = ex2+y3

4. z = ex2y3

MATH 200

EXAMPLE 1

f(x, y) = 3x2 � 2y + 1

fx(x, y) = 3(2x1) � 0 + 0

= 6x

f(x, y) = 3x2 � 2y + 1

fy(x, y) = 0 � 2(1) + 0

= �2

WITH RESPECT TO X, THESE ARE CONSTANT TERMSPOWER RULE

WITH RESPECT TO Y, THESE ARE CONSTANT TERMS

POWER RULE

MATH 200

EXAMPLE 2g(x, y) = xy2 � ln y

gx(x, y) = (1)y2 � 0

= y2

g(x, y) = xy2 � ln y

gy(x, y) = x(2y) � 1

y

= 2xy � 1

y

WITH RESPECT TO X, THIS WHOLE TERM IS

CONSTANT

y2 IS TREATED AS CONSTANT HERE, SO IT’S LIKE DIFFERENTIATING 5x

WITH RESPECT TO xx IS TREATED AS CONSTANT

HERE, SO IT’S LIKE DIFFERENTIATING 5y2 WITH

RESPECT TO y

MATH 200

EXAMPLE 3

h(x, y) = ex2+y2

hx(x, y) = 2xex2+y2

h(x, y) = ex2+y2

hy(x, y) = 2yex2+y2

THE DERIVATIVE OF eu(x) IS u’(x)eu(x), SO IN

TERMS OF PARTIAL DERIVATIVES, WE SHOULD WRITE

ux(x,y)eu(x,y)

THE DERIVATIVE OF eu(x) IS u’(x)eu(x), SO IN

TERMS OF PARTIAL DERIVATIVES, WE SHOULD WRITE

uy(x,y)eu(x,y)

MATH 200

EXAMPLE 4

z = ex2y3

�z

�x= 2xy3ex2y3

z = ex2y3

�z

�y= 3x2y2ex2y3

∂

∂x(x2y3) = 2xy3

WE CAN WRITE “THE PARTIAL DERIVATIVE WITH RESPECT TO

x OF x2y3” LIKE THIS:

MATH 200

WHAT PARTIAL DERIVATIVES GIVE US▸ Let’s look at f(x,y) = 3x2 - 2y + 1 from Example 2 at the

point (1,2)

▸ We found that fx(x,y) = 6x

▸ Evaluating the x partial at (1,2) we get fx(1,2) = 6(1) = 6

▸ What does this 6 tell us?

▸ The rate of change of f (or z) in the x-direction at (1,2) is 6

▸ The slope of the line tangent to the trace of f(x,y) on the plane y=2 is 6

MATH 200

∂z

∂x

!!!!(1,0)

= 6 =⇒ ⟨1, 0, 6⟩

L :

⎧⎪⎨

⎪⎩

x = 1 + t

y = 2

z = 6t

MATH 200

HIGHER ORDER PARTIAL DERIVATIVES▸ Consider the function z = 3x2 - x3y4

▸ Let’s find the two first order partial derivatives:∂z

∂x= 6x− 3x2y4;

∂z

∂y= 4x3y3

▸ We could now differentiate either of these with respect to x or with respect to y…

▸ …for a total of four second order partial derivatives ∂

∂x

!∂z

∂x

"=

∂2z

∂x2;

∂

∂y

!∂z

∂x

"=

∂2z

∂y∂x;

∂

∂x

!∂z

∂y

"=

∂2z

∂x∂y

∂

∂y

!∂z

∂y

"=

∂2z

∂y2;

MATH 200

∂2z

∂x2= 6− 6xy4

∂2z

∂y∂x= 12x2y3

∂2z

∂x∂y= 12x2y3

∂2z

∂y2= 12x3y2

NOTICE THAT THE SECOND ORDER MIXED PARTIALS ARE THE SAME!

THIS WILL BE TRUE FOR ANY TIME THEY

ARE BOTH CONTINUOUS

MATH 200

IMPLICIT DIFFERENTIATION▸ Recall from calc 1:

x4 + y4 = xy

d

dx(x4 + y4) =

d

dx(xy)

4x3 + 4y3 = y + xdy

dxdy

dx=

4x3 + 4y3 � y

x

TREAT Y AS AN IMPLICIT

FUNCTION OF X

MATH 200

IMPLICIT DIFFERENTIATION WITH THREE VARIABLES

Find�z

�xxey+z � 2z2 = 3y + 1

�

�x(xey+z � 2z2) =

�

�x(3y + 1)

ey+z + x�z

�xey+z � 4z

�z

�x= 0

∂z

∂x=

−ey+z

xey+z − 4z

TREAT Z AS A FUNCTION OF X

AND Y AS A CONSTANT

MATH 200

ONE MORE EXAMPLE▸ Compute dy/dz for 3xy2 � zey = 4z3

�

�z(3xy2 � zey) =

�

�z(4z3)

3x

�2y

�y

�z

��

�ey + zey �y

�z

�= 12z2

6xy�y

�z� ey � zey �y

�z= 12z2

�y

�z(6xy � zey) = 12z2 + ey

�y

�z=

12z2 + ey

6xy � zey

DIFFERENTIATE BOTH SIDES WITH RESPECT TO Z, TREATING Y AS A

FUNCTION OF Z AND X AS A CONSTANT

PRODUCT RULE

GET ALL OF THE DY/DZ TERMS ON ONE SIDE AND FACTOR

SOLVE

MATH 200

RECAP▸ To compute the partial derivative of f

with respect to x, we…

▸ Treat all other variables as constants

▸ Use all of the derivative rules we know from Calculus 1

▸ E.g. f(x,y) = x3+y3

▸ fx(x,y) = 3x2

▸ When we evaluate partial derivative of f with respect to x at a point (x0,y0), we get the slope of the line tangent to the trace of f on the plane y = y0

(x0,y0)

Trace on y=y0