9. The differential of the function. Higher order derivatives. Leibniz’s formula.

Answer:

Definition. We say that f is differentiable at point x₀ (a;b), if : f(x₀) = A(x-x₀) + ō(x-x₀)

f(x₀) is increment of function, A(x-x₀) – f’(x₀) x is increment, ō(x-x₀) is infinitesimal.

From (1) => dy = f’(x₀)dx (2) the symbols “dy” and “dx” are called differentials.

If dx 0 , then we can divide both sides of (2) by dx to obtain = f’(x₀) (3)

Formula (2) is said to express (3) in differential form.

Differential formulas:

1) d(c) = 0

2) d(cf) = cdf

3) d(f+g) = df + dg

4) d(fg) = (df)g + f(dg)

5) d( ) = (df)g – (dg)f/g²

Leibniz’s formula:

(fg)^(k) = ^m_k f^(m)g^(k-m) =1fg^(k) + kf’g^(k-1) + f’’g^(k-2) + …+ 1f^(k)g.

C^m_k = ; C⁰_k = =1 ; C¹_k = = k .

10.L’Hospital’srule.The other indeterminate form.

Answer:

Theorem. Suppose that f and g are differentiable and g’ has no zeros on (a;b)

(x)= (x)=0 (1) or

(x)= (x)= (2)

And suppose that =L (3) Then =L (4)

Proof:

Suppose,that ε>0. From (3), there is x₀ (a;b) such that | – L |<ε if x₀<c<b(5)

Cauch’s Theorem implies that if x and t are in [x₀,b) , then for every c between them , and therefore in (x₀,b) , such that

(g(x)-g(t))f’(c)=(f(x)-f(t)g’(c)) (6)

Since g’ has no zeros in (a,b) Lagranzh’s Theorem implies that g(x)-g(t) 0 if x,t (а,b)

This means that g cannot have more than one zero in (a,b). Therefore, we can choose x₀ so that , in addition to (5), g has zero in [x₀,b)

Then (6) can be rewritten as = ,

so implies that as | -L|<ε if x,t (a,b) (7)

If (1) holds, Let x be fixed in [x₀,b) , and consider the function

10.2 G(t)= -L

From (1) (t)= (t)=0

So (t)= -L (8)

Since, І G(t) І<ε if t (x₀,b) because of (7) ,(8) implies that І –L| ε.This holds for all x in (x₀,b) , which implies (4)

L’Hospital’s rule используется : (0/0) and (&/&).The other indeterminate form:0&, &-&, 0⁰,1^&,&⁰. Ары каратай далелдеу керек еще!!!!»

1. The indeterminate form 0 : we say that a product fg is of the form 0 , as x->b,if of the factors approaches 0 and the order approaches as x->b- In this case, it may be useful to apply L’Hospital’s rule after writing f(x)g(x)= =( ), f(x)g(x)= (since one of the rations is the form ( ) and the other is of the form ( ))

Similar statements apply to limits as x->b+, x->b, x->

b)The indeterminate form : A difference (f-g) is of the form ( ) as x->b, if

(x)= (x)=

In this case reduced to a common dehominator.

c)The indeterminate form 0⁰,1^&, ⁰. In this three cases, we use formula U^V=e^VLnU (f(x))^g(x)=e^g(x)Lnf(x).