Suppose f is a two variable real valued function, and we’d like to know the rate
of change of f at (x0 , y0 ) in some arbitrary direction, say v = ha, bi1. This is
accomplished by computing
d
.
f(x(t), y(t))
dt
t=0
In order to effectively compute the derivative of this composition of functions, we
need the multi-variable chain rule.
Theorem 1. If f(x, y) is differentiable at (x(t0 ), y(t0 )), and x(t) and y(t) are
differentiable at t0 , then f(x(t), y(t)) is differentiable at t0 , in fact,
d
dx
∂f
dy
∂f
f(x(t), y(t))
(x(t0 ), y(t0 )) ·
(t0 ) +
(x(t0 ), y(t0 )) ·
(t0 ).
=
dt
∂x
dt
∂y
dt
t=t0
More briefly, we have
df
∂f dx
∂f dy
=
·
+
·
.
dt
∂x dt ∂y dt
Proof. (sketch) This more or less follows from composing the linearization (or equation of the tangent plane) of f at (x(t0 ), y(t0 )) with the linearizations (or equations
of the tangent lines) of x(t) and y(t) at t = t0 . In the end, we essentially get
∂f dy
∂f dx
·
+
·
(t0 )(t − t0 ).
f(x(t), y(t)) ≈ f(x(t0 ), y(t0 )) +
∂x dt ∂y dt
If the expression on the right is the tangent line to f(x(t), y(t)) at t = t0 , then the
coefficient on the term (t − t0 ) must be the derivative of f(x(t), y(t)) evaluated at
t = t0 ).
This generalizes to the following.
Theorem 2. Suppose f is a real valued function in n variables. Suppose also
that x1 , . . . , xn are all real valued function in the variables u1 , . . . , um .
Then
f(x1 , . . . , xn ) is a function in the variables u1 , . . . , um . If x1 , . . . , xn are all differentiable at (u1 , . . . , um ), and f is differentiable at (x1 (u1 , . . . , um ), . . . , xn (u1 , . . . , um )),
then f(x1 , . . . , xn ) is differentiable at (u1 , . . . , um ), in fact, for any 1 6 j 6 m
n
X
∂f
∂f ∂x1
∂f ∂x2
∂f ∂xn
∂f ∂xk
=
·
+
·
+ ··· +
·
=
·
.
∂uj
∂x1 ∂uj ∂x2 ∂uj
∂xn ∂uj
∂xk ∂uj
k=1
Example 1. Let f(x, y) = x2 + y2 and x = 2t + 1 and y = 1 + t. Then f(x, y) is a
function of t. Compute the derivative of f with respect to t at t = 0.
1Next time we’ll see that, perhaps, it’s best to choose v to be a unit vector
1
2
Solution. We use the chain rule. Note that fx = 2x and fy = 2y, also, x 0 (t) = 2
and y 0 (t) = 1. So
df
(0)
dt
∂f
∂f
(x(0), y(0)) · x 0 (0) +
(x(0), y(0)) · y 0 (0)
∂x
∂y
= 2(x(0)) · 2 + 2(y(0)) · 1
=
= 4+2
= 6.
Problem 1. Let f(x, y, z) = sin(xyz) and x = 2u + 3v, y = arctan(uv) and
z=
u2
v2 +1 .
Compute the first partial derivatives of f with respect to u and v.