Maths3

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths3

The notion of derivative, rules of dierentiation
3.1 Basic notions

Let f be a function dened on some interval and suppose that x0 is a point inside this interval.

Denition 3.1 We say that f is dierentiable at the point x0 if the limit
h0

lim

f (x0 + h) - f (x0 ) h

f (x0 ).

exists and is nite. This limit is called the derivative of f at x0 and is denoted by The function f is said to be dierentiable on an interval, if it is dierentiable at each interior point of the interval.

The quotient occurring in the above limit is called the dierence quotient for f at x0 .

Example 3.1 Consider the function f (x) = x2 . The dierence quotient at x0 is
(x0 + h)2 - x2 f (x0 + h) - f (x0 ) 0 = = 2x0 + h h h

which approaches 2x0 as h 0. Consequently
f (x0 ) = 2x0 .

Analogously, it can be shown that for f (x) = xn where n is a natural number we have
f (x0 ) = nxn-1 . 0

Example 3.2 Consider the function f (x) = |x| and examine its dierence quotient at x0 = 0. It is clear that
f (h) - f (0) |h| = = h h +1, 1 -1

if h > 0 . if h < 0

The limit of this quotient as h 0 does not exist, since the right-hand limit equals while the left-hand limit is -1. Thus the function f is not dierentiable at 0. However, it is dierentiable at all other points, namely we have f (x) = 1 if x > 0 and f (x) = -1 if x < 0.
Theorem 3.1 If f is dierentiable at x0 , then it is also continuous there.
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3.2.

TANGENTS OF CURVES

Indeed, for any sequence hn 0 the dierence quotient has a nite limit, therefore it is bounded, that is
f (x0 + hn ) - f (x0 ) K hn

for some number K > 0. This means that
|f (x0 + hn ) - f (x0 )| K|hn | ,

hence f (x0 + hn ) f (x0 ) if n . Therefore, by denition, we have
xx0

lim f (x) = f (x0 ) .

Thus, f is continuous at x0 . The reverse of this statement is not true.In the previous example it was shown that the function f (x) = |x| is not dierentiable at 0, however it is continuous everywhere, in particular at 0.
3.2 Tangents of curves

The geometric interpretation shows that the derivative f (x0 ) equals to the slope of the tangent line to the graph of f at x0 . Using this we can determine the equation of the tangent line to the graph of a dierentiable function drawn at the point x0 :
y = f (x0 )(x - x0 ) + f (x0 ) .

For example, the equation of the tangent line to the graph of f (x) = x3 at x0 = 1 is
y = 3(x - 1) + 1,

the equation of the tangent line to the graph of f (x) = sin x at x0 = 0 is y = x, since
f (0) = lim
h0

properties of limits we get the following rules.

f (h) - f (0) sin h = lim =1. h0 h h Let us assume that the functions f and g are dierentiable at x. From the

Derivative of sums and scalar multiples of functions If and
trary real numbers, then f + g is also dierentiable at x and
(f (x) + g(x)) = f (x) + g (x) ,

are arbi-

Derivative of products f · g is dierentiable and
(f (x) · g(x)) = f (x) · g(x) + f (x) · g (x) ,

Derivative of quotients if g(x) = 0, then f /g is dierentiable and
f (x) g(x) = f (x)g(x) - f (x)g (x) . g(x)2

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3.3.

HIGHER ORDER DERIVATIVES

As an example we prove the rule of dierentiation of product of functions.
f (x + h) · g(x + h) - f (x) · g(x) = h f (x + h) · g(x + h) - f (x + h) · g(x) + h f (x + h) · g(x) - f (x) · g(x) = h f (x + h) g(x + h) - g(x) f (x + h) - f (x) + g(x) h h

Because of the continuity of f we have f (x + h) f (x) if h 0, thus the limit of the dierence quotient is
f (x)g (x) + f (x)g(x)

as h 0.

Example 3.3 Show that the tangent line drawn at any point to the graph of the function f (x) = 1/x encloses with the coordinate-axis the same area. Obviously, it is enough to consider points x0 > 0. Since f (x0 ) = -1/x2 , 0 therefore the equation of the tangent line to the graph at x0 is
y=- 1 1 2 (x - x0 ) + x . x0 0

The intercepts of this line are if x = 0 then on the y-axis: b =
2 x0

if y = 0 then on the x-axis: a = 2x0 Thus, the area of the triangle enclosed is
T = 2 1 =2 · 2x0 · 2 x0

which is independent of the point x0 .
Exercise 3.1 Find the derivative of the following functions:
y = 5x3 - 4x2 + 3, y= 2x + 1 , x10 1 y= , x y = ex sin x, tan x , y= 1 + tan2 x x2 - x , 4 2 1x -1 y= , 2 x2 + 1 1 3 y= x, x y = ln x · cos x, y= x-3 , x-5 y = x, y= 1 y =x+ , x 3 y = x2 , y = x sin x, y = ln 10x ,

y = (x2 + 1)ex , y = sin 4x,

3.3

Higher order derivatives

If a function f is dierentiable at each point of an interval then the mapping x f (x) is called the derivative function of f . If f is dierentiable at a point x0 , then we say that f is dierentiable twice there. In this case we use the notation
f (x0 )

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3.4.

HOMEWORK:

instead of (f ) (x0 ) and we call f (x0 ) the second derivative of f at x0 . Similarly we dene the n-th derivative of f at x0 denoted by
f (n) (x0 )

For example for the function f (x) = 1/x we have
f (n) (x0 ) = (-1)n n! xn+1

for all x0 = 0 and for all natural numbers n 1.

Exercise 3.2 Determine 1) y = x5 - 2x3 + 1 2) y = x3 - 8x + 2 3) y = 4 x3 4) y = x4 ln x 5) y = 25x

the indicated higher derivative of the following functions:
y (5) =? y (5) =? y =? y =? y =? y =?
(5)

6) y = x-1 7) y = x2 e2x 8) y = ex sin x
3.4

x3

y =? y (4) =?

Homework:

· Book ch6.7, ch6.9 · Practice the rules of direntiation (sum, products and quotient)! · Exercise given above + the exercise in ch6.7, ch6.9

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