Calculus III:

Differentials
Derivatives of functions of more
than one variable

1. Defnitions and rules

We had differentials for functions of one variable. We also have them for functions of more than one variable. Between them, there are some small differences.

Given the function f(x,y), the differential df is given by:

The extension to functions of three or more variables, for example f(x,y,z) is given by

2. Examples

Example 1

Let's consider f(x,y) = xy

The differential for this function is:

df = ydx + xdy

Example 2

Let's consider a three variables function f(x,y,z) = x²y²z²

The differential for this function is:

df = 2xy²z²dx + 2x²yz²dy + 2x²y²zdz

3. Using chain rule

3.1. functions of one variable

Recall that the standard chain rule for functions of one variable :

If h(x) = f(g(x)), then

We can also have:

y = f(x) and x = g(t). Then, the chain rule is written as:

dy/dt = dy/dx . dx/dt

There are many different formulas depending upon the number of variables that a function has.

3.2. function of two variables of one variable

z = f(x,y) , x = g(t) , y = h(t)

If we substitute in z = f(x,y) for x and y, z will be a function
of t only .

The chain rule for this case is:

dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt)

We differentiate f with respect to each variable x and y in it and then multiplying each of these by the ordinary derivative of that variable with respect to t. Therefore, we add all this up.

Example 1:

f(x,y) = exp{xy}, x = t², y = 2t

dz/dt = ∂f/∂x . dx/dt + ∂f/∂y . dy/dt

dz/dt = (y exp{xy}) . (2t) + (x exp{xy}) . (2) =

2 t y exp{xy} + 2x exp{xy}

Substituting in for x and y gives:

dz/dt = 2 t (2t) exp{2t³} + 2t² exp{2t³} =
4t² exp{2t³} + 2t² exp{2t³} =

dz/dt = 6t² exp{2t³}

We can substitute directly t in z, derive ordinarily and get the same result:

f(x,y) = exp{xy} = exp{2t³}

dz/dt = 6t² exp{2t³}

Example 2:

z = f(x,y) , y = g(x)

The chain rule for dz/dx is:

dz/dx = ∂f/∂x . dx/dx + ∂f/∂y . dy/dx
= ∂f/∂x + ∂f/∂y . dy/dx

z = xy³ , y = sin(x)

dz/dx = ∂f/∂x + ∂f/∂y . dy/dx = y³ + (3xy² ) . (cos(x))

dz/dx = y³ + 3 x cos(x) y²

3.3. function of two variables of two variables

z = f(x,y) , x = g(s,t) , y = h(s,t)

Inthis case we cannot have an ordinary derivative dz/dx, or dz/dg, or dz/ds, or dz/dt.

If we substitute in for x and y we would get that z is a function of the two variables s and t . The chain rule for both of these cases are:

∂z/∂s = ∂f/∂x . ∂x/∂s + ∂f/∂y . ∂y/∂s

∂z/∂t = ∂f/∂x . ∂x/∂t + ∂f/∂y . ∂y/∂t

3.4. General case:
function of n variables of m variables

Let z is a function f of n variables x1, x2, x3, ..., xn , and that each of these variables are in turn functions of m variables, t1, t2, t3, ..., tm.

Then for any variable ti, we have the following chain rule:

∂z/∂ti = ∂f/∂x1 . ∂x1/∂ti + ∂f/∂x2 . ∂x2/∂ti + ... + ∂f/∂xn . ∂xn/∂ti

Example 1

w = f(x, y, z) x = g1(t), y = g2(t), z = g3(t)

We compute dw/dt:

dw/dt = (∂f/∂x )(∂x/∂t) + (∂f/∂y )(∂y/∂t) + (∂f/∂z )( ∂z/∂t)

Example 2

w = f(x, y, z) x = g1(r, s, t), y = g2(r, s, t), z = g3(r, s, t)

We compute ∂w/∂r, ∂w/∂s, and ∂w/∂t:

∂w/∂r = (∂f/∂x)(∂x/∂r) + (∂f/∂y)(∂y/∂r) + (∂f/∂z)(∂z/∂r)

∂w/∂s = (∂f/∂x)(∂x/∂rs) + (∂f/∂y)(∂y/∂s) + (∂f/∂z )( ∂z/∂s)

∂w/∂t = (∂f/∂x)(∂x/∂t) + (∂f/∂y )( ∂y/∂t) + (∂f/∂z )( ∂z/∂t)

Example 3

z = f(x,y), x = r cos (θ), y = r sin (θ)

If r varies, we compute ∂f/∂θ. If r is constant, we compute df/dθ. We suppose that r is constant and ask to compute : d²f/dθ².

df/dθ = (∂f/∂x)(dx/dθ) + (∂f/∂y)(dy/dθ)
= - r sin (θ) (∂f/∂x) + r cos (θ)(∂f/∂y)

d²f/dθ² = ∂(∂f/∂θ)/∂θ =
∂(∂[- r sin (θ) (∂f/∂x) + r cos (θ)(∂f/∂y)]/∂θ)/∂θ

∂f/∂x and ∂f/∂y are both functions of x and y which are in turn functions of θ. So

∂(∂f/∂x)/∂θ =
- r sin (θ) ∂(∂f/∂x)/∂x; + r cos (θ) ∂(∂f/∂x)/∂y =
- r sinθ ∂²f/∂x² + r cos (θ) ∂²f/∂x)∂y

∂(∂f/∂y)/∂θ =
- r sin (θ) ∂²f/∂x∂y + r cos (θ) ∂²f/∂y²

Rearranging and combining leads to :

d²f/dθ² = - r cos (θ) (∂f/∂x) - r sin (θ) (∂f/∂y)
+ r² sin² θ (∂²f/∂x²) - 2 r² sin (θ) cos (θ) (∂²f/∂y ∂x)
+ r² cos²(θ) (∂²f/∂y²)

Example 4: Implicit differentiation

Let a function of one variable in the form f(x, y) = 0,
and compute dy/dx.

Differentiating both sides with respect to x, yields:

f_x(dx/dx) + f_y(dy/dx) = 0. Hence:

∂y/∂x = - f_x/f_y

Exemple:

Find dy/dx for 2 x y + y = 3 - 4y

2 x y + y = 3 - 4y → 2 x y + 5y - 3 = 0

dy/dx = - f_x/f_y = - 2 y/(2x+ 5)

dy/dx = - 2 y/(2x+ 5)

With three variables, let's consider the fonction z = f(x, y) that tkes the form f(x, y, z) = 0.

We want to find

∂f/∂x and ∂f/∂y

• ∂f/∂x

Differentiating both sides with respect to x, yields:

(∂f/∂x)(∂x/∂x) + (∂f/∂y)(∂y/∂x) + (∂f/∂z)(∂z/∂x) = 0

We have: ∂x/∂x = 1 and ∂y/∂x = 0. It remains:

∂f/∂x + (∂f/∂z)(∂z/∂x) = 0

Hence:

∂z/∂x = - ∂f/∂x/(∂f/∂z) . Or

∂z/∂x = - f_x/f_z

• ∂f/∂y

Differentiating both sides with respect to y, yields:

(∂f/∂x)(∂x/∂y) + (∂f/∂y)(∂y/∂y) + (∂f/∂z)(∂z/∂y) = 0

We have: ∂y/∂y = 1 and ∂x/∂y = 0. It remains:

∂f/∂y + (∂f/∂z)(∂z/∂y) = 0

Hence:

∂z/∂y = - ∂f/∂y/(∂f/∂z) . Or

∂z/∂y = - f_y/f_z