Partial Derivatives

PARTIAL DERIVATIVES

Functions of two or more independent variables appear in many practical problems more often than functions of one independent variable. The concept of derivative of a single variable function f(x) is extended to functions of two or more variables. Suppose f(x, y) is a function of two independent variables x and y, we treat y as constant and find the derivative of f(x, y) w.r.to x, then the derivative is called a partial derivative. Partial derivatives find applications in a wide variety of fields like fluid dynamics, electricity, physical sciences, econometrics, probability theory etc.

Definition 3.2.1 Let z = f(x, y) be a real function of two independent variables x and y. Let (x0, y0) be a point in the domain of f. The partial derivative of f(x, y) w.r.to x at (x0, y0) is the limit.

if the limit exists. Then it is denoted by

Similarly, partial derivative of f(x, y) w.r.to y at (x0, y0) is the limit

If the partial derivatives of z = f(x, y) exist at any point in its domain, then the partial derivatives w.r.to x is simply written as as assuming the point (x, y).

Similarly, the partial derivative w.r.to y is written as

1. Geometrical meaning of

Let z = f(x, y) be a real function of two independent variables x, y. For every point (x, y) in its domain R in the xy plane there is a real number z, where f(x, y) = z. The set of all points (x, y, z), where z = f(x, y), in space determine a surface S. This surface is called the graph of the function f. Thus z = f(x, y) represents a surface in space.

The equation y = y0 represents a vertical plane (parallel to xoz plane) 

intersecting the surface in a curve C : z = f(x, y0). The partial derivative at (x0, y0). represents the slope of the tangent to this curve at the point (x0, y0, z0), where z0 = f(x0, y0).

Similarly, the partial derivative at (x0, y0) is the slope of the tangent to the curve z = f(x0, y) at the point (x0, y0, z0), where z0 = f(x0, y0)·

Note

(1) If z = f(x, y) then f(x, y) = k for all points in the domain of f, where k is a constant, is called a level curve of function f.

(2) If u = f(x, y, z) be a function of three independent variables x, y, z then the graph of ƒ is a 4-dimensional surface.

f(x, y, z) = c, where c is a constant, is called a level surface. For different c, we have different level surfaces.

2. Partial derivatives of higher order

Let z = f(x, y) be a function of 2 independent variables. The derivatives are called partial derivatives of first order, which are again functions of x, y and can be differentiated partially w.r.to x, y. These are called partial derivatives of second order and denoted by

It can be shown that if fx, fy, and fxy, are continuous then fxy = fyx. In fact the elementary functions that we come across satisfy these conditions. In many practical applications also these conditions are satisfied. So, we shall assume this in our discussions.

Differentiating the second order derivatives partially w.r.to x, y we get third order derivatives.

3. Homogeneous functions and Euler's theorem

Definition 3.2.2 A function f(x, y) is said to be homogeneous of degree (or order) n if f(tx, ty) = tn f(x, y) for any positive t.

For example (1): f(x, y) = is homogeneous of degree 2.

Theorem 3.2.1 Euler's theorem

If f(x, y) is a homogeneous function of degree n in x and y having continuous partial derivatives, then

Proof Given f(x, y) is a homogeneous function in x and y of degree n, we can write

This theorem can be extended to homogeneous function of any number of variables. If f(x, y, z) is a homogeneous function of degree n in three independent variables x, y, z and differentiable then

WORKED EXAMPLES

Example 1 

If u = log (x3 + y3 + z3 — 3xyz), then prove that

Solution 

Given u = log (x3 + y3 + z3 – 3xyz)

Example 2 

If u = (x − y)4 + (y − z)4 + (z − x)4, then find the value of

Solution 

Given u = (x − y)4 + (y − z)4 + (z − x)4

Example 3 

If u = (x – y) (y – z) (z – x) then prove that

(i)

Solution 

Given u = (x − y) (y - z) (z - x)

(ii) u is a homogeneous function of degree 3, since

u (xt, yt, zt) = (tx - ty) (ty - tz) (tz - tx) = t3 (x − y) (y − z) (z − x)

= t3 (x - y) (y - z) (z − x)

= t3u(x, y, z)

So, by Euler's theorem, we get

Example 4 

Solution 

⸫ ƒ is a homogeneous function of degree 3 in x, y

Example 5 

Solution 

⸫ u is a homogeneous function of degree 0 in x, y, z

By Euler's theorem, we get

Example 6 

Solution 

Given u is a function of x, y, z

⸫ u(x, y, z) is a homogeneous function of degree 0 in x, y, z

By Euler's theorem, we get

Example 7 

Solution 

Given u = f(x - y, y - z, z - x)

Put x1 = x - y, y1 = y - z, z1 = z - x

then u = f(x1, y1, z1), where x1, y1, z1 are functions of x, y, z.

Example 8 

Solution 

Example 9 

Solution 

f(x, y) is a homogeneous function of degree 2

So, by Euler's theorem on homogeneous functions, we get

Another result on homogeneous functions which follow from Euler's theorem is given below.

Theorem 3.2.2 If u(x, y) is homogeneous function of degree n in x and y with all first and second derivatives continuous, then

Proof Given u(x, y) is a homogeneous function of x and y of degree n. So, by Euler's theorem

Differentiating (1) partially w.r.to x, we get

Now differentiating (1) w.r.to y, we get

Example 10 

Solution 

Example 11 

Solution 

= t2u(x, y)

⸫ u(x, y) is homogeneous of degree 2 and it is differentiable twice, partially. 

⸫ by theorem 3.2.2,

Example 12 

Solution 

Example 13 

If u = xy, then show that (i) uxy = uyx

(ii) uxxy = uxyx

Solution 

Given u = xy

From (3) and (6), we get (ii)     ⸫ uxxy = uxyx

Example 14 

Solution 

Example 15 

Solution 

Given   r2 = x2 + y2 + z2

4. Total derivatives

Let u = f(x, y) be a function of 2 variables x, y. If x and y are continuous function of t then z will be ultimately a function of t only or z is a composite function of t. Then we can find the ordinary derivative du/dt which is called the total derivative of u to distinguish it from the partial derivatives

We have

Proof u = f(x, y), x = F(t), y = G(t).

Giving increment Δt to t will result in increments

Δx, Δy and Δu in x, y and u.

Cor (1): In differential form the result (1) can be written as

du is called the total differential of u.

Similarly, if u = f(x, y, z) of 3 independent variable x, y, z then the total differential is

Cor (2): Change of variables

If u = f(x, y) where x and y are function of t1, t2 then u is ultimately a function of t1, t2 and so z is a composite function of t1, t2.

Then we have partial derivatives of u w.r.to t1, t2

Cor (3): Differentiation of implicit functions

The equation f(x, y) = 0 defines y implicitly as a function of x. Suppose the function f(x, y) is differentiable, then the total differential

WORKED EXAMPLES

Example 1 

Solution 

Given u = x2 y3, x = log t, y = et 

So u is ultimately a function of t

Example 2 

Solution 

Example 3 

Solution 

Given u = sin-1(x - y), where x = 3t, y = 4t3

Example 4 

Solution 

We have x = et,      y = t2

Example 5 

Solution 

Given u = x2y,  x = t2, y = et

Example 6 

Solution 

Given u = cos (x2 + y2) and a2x2 + b2y2 = c2

Example 7 

Solution 

Given z is a composite function of u and v.

⸫ we have partial derivatives of z w.r.to u and v

and x = eu + e-v and y = e-u - ev

Example 8 

Solution 

Since x and y are functions of r and θ, u is a composite function of r and θ. So we have partial derivatives of u w.r.to r, θ

Example 9 

Solution 

Given z = f(x, y) and x = eu cos v, y = eu sin v

⸫ z is a composite function of u and v.

Example 10 

Solution 

Example 11 

Solution 

u and v can be considered as functions of x and y.

We have u2 + 2v2 = 1 − x2 + y2

Differentiating partially w.r.to x, we get

Differentiating partially w.r.to x, we get

Example 12 

Solution 

Given z = where x3 + y3 + 3axy = 5a2 is an implicit relation and y is implicitly a function of x.

So, z is composite function of x.

Given x3 + y3 + 3axy = 5a2

Differentiating w.r.to x, we get

Example 13 

If z is a function of x and y and x = u cos α - v sin α, y = u sin α + v cos α, then show that

Solution 

Given z is a composite function of u and v

and x = u cos α - v sin α, y = u sin α + v cos α

Example 14 

If z = f(x, y) where x = u2 - v2, y = 2uv, prove that

Solution 

Given z = f(x, y) where x = u2 - v2, y = 2uv

Since z is a function of x, y,

But x and y are function of u and v

So z is ultimately a function of u and v.

But x = u2 - v2, y = 2uv

Note: The same problem is asked as below also. If where

Example 15 

Transform equation zxx + 2zxy + zyy = 0 by changing the independent variables using u = x - y and v = x + y

Solution

Given zxx + 2zxy + zyy = 0

and u = x – y,   v = x + y

In the given equation independent variables are x and y.

We have to change them to u and v. 

So, we treat z as a function of u and v

Substituting (2), (3) and (4) in (1), we get

EXERCISE 3.2

ANSWERS TO EXERCISE 3.2