Partial Derivatives: Definition, Rules, Interpretations, Higher Derivatives, Schemes and Mind Maps of Mathematics

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PARTIAL DERIVATIVES

ELECTRONIC VERSION OF LECTURE

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

OUTLINE

(^1) PARTIAL DERIVATIVES

(^2) DIFFERENTIALS

(^3) MATLAB

Partial derivatives Definition

Partial derivatives Definition

EXAMPLE 1.

If f (x, y) = x^3 + x^2 y^3 − 2 y^2 , find f x′ (2, 1) và f y′ (2, 1)

SOLUTION. Holding y = 1, we receive

g (x) = f (x, 1) = x^3 + x^2 − 2 ⇒ g ′(x) = 3 x^2 + 2 x

⇒ g ′(2) = 3 × 22 + 2 × 2 = 16 = f (^) x′ (2, 1).

Holding x = 2, we receive

h(y) = f (2, y) = 23 + 22 y^3 − 2 y^2 ⇒ h′(y) = 12 y^2 − 4 y ⇒ h′(1) = 12 × 12 − 4 × 1 = 8 = f (^) y′ (2, 1).

Partial derivatives Rule for Finding Partial Derivatives of z = f (x, y)

EXAMPLE 1.

If f (x, y) = arctan

x y

, find f x′ and f y′.

SOLUTION.

f (^) x′ = ∂^ f ∂ x

³ (^) x y

´ 2 ·^

y = y x^2 + y^2

f (^) y′ = ∂ f ∂ y

³ (^) x y

μ − x y^2

x x^2 + y^2

Partial derivatives Interpretations of Partial Derivatives

1 f x′ (x 0 , y 0 ) is the slope of the tangent lines T 1 at

P (x 0 , y 0 , f (x 0 , y 0 )) to the trace C 1 of S in the plane

y = y 0.

2 f y′ (x 0 , y 0 ) is the slope of the tangent lines T 2 at

P (x 0 , y 0 , f (x 0 , y 0 )) to the trace C 2 of S in the plane

x = x 0.

Partial derivatives Interpretations of Partial Derivatives

C 2 : z = 3 − 2 y^2 , x = 1. The slope of the tangent line T 2

to parabola C 2 at (1, 1, 1) is f y′ (1, 1) = −4.

Partial derivatives Higher Derivatives

SECOND PARTIAL DERIVATIVES ¡ f (^) x′

x =^

∂ x

μ ∂ f ∂ x

∂^2 f ∂ x^2 = f (^) xx′′

¡ f (^) x′

y =^

∂ y

μ ∂ f ∂ x

(^2) f ∂ x ∂ y = f (^) x y′′

³ f (^) y′

x =^

∂ x

μ ∂ f ∂ y

∂^2 f ∂ y ∂ x = f (^) y x′′ ³ f (^) y′

y =^

∂ y

μ ∂ f ∂ y

∂^2 f ∂ y^2 =^ f^ y y′′

Partial derivatives Higher Derivatives

EXAMPLE 1.

Find the second partial derivatives of f (x, y) = x ye y^.

SOLUTION. We have

f (^) x′ = ye y^ ; f (^) xx′′ = 0; f (^) x y′′ = e y^ + ye y^ ; f (^) y′ = x(e y^ + ye y^ ); f (^) y y′′ = x(2e y^ + ye y^ ); f (^) y x′′ = e y^ + ye y^.

Partial derivatives Higher Derivatives

EXAMPLE 1.

If f (x, y) = sin(x y), find ∂

(^2) f ∂ y^2

(^3) f ∂ y^2 ∂ x

SOLUTION. We have

∂ f ∂ y = x cos(x y); ∂

(^2) f ∂ y^2 = −x^2 sin(x y).

∂^3 f ∂ y^2 ∂ x

∂ x

− x^2 sin(x y)

= − 2 x sin(x y) − x^2 y cos(x y).

Partial derivatives Partial Differential Equations

EXAMPLE 1.

Show that the function u(x, y) = ex^ sin y is a solution of

Laplace’s equation.

SOLUTION. We have

u′ x = ex^ sin y, u′ y = ex^ cos y, u′′ xx = ex^ sin y, u′′ y y = −ex^ sin y.

So u′′ xx + u′′ y y = ex^ sin y − ex^ sin y = 0. Therefore

u(x, y) = ex^ sin y satisfies Laplace’s equation.

Partial derivatives Partial Differential Equations

THE WAVE EQUATION

∂^2 u ∂ t 2 =^ a

2 ∂^2 u ∂ x^2

describes the motion of a waveform, which could be

an ocean wave, a sound wave, a light wave, or a wave

traveling along a vibrating string.

Differentials Definition

Consider function z = f (x, y), and suppose x changes

from x 0 to x 0 + ∆x, and y changes from y 0 to y 0 + ∆y.

Then the corresponding increment of z is

∆z = f (x 0 + ∆x, y 0 + ∆y) − f (x 0 , y 0 ).

Differentials Definition

DEFINITION 2.

If z = f (x, y), then f is differentiable at (x 0 , y 0 ) if

∆z = f (x 0 + ∆x, y 0 + ∆y) − f (x 0 , y 0 ) can be expressed in

the form

f (x 0 + ∆x, y 0 + ∆y) − f (x 0 , y 0 ) = ∆z = = f (^) x′ (x 0 , y 0 )∆x + f (^) y′ (x 0 , y 0 )∆y + ε 1 ∆x + ε 2 ∆y,

where ε 1 , ε 2 → 0 as (∆x, ∆y) → (0, 0).