Unit – 1 | unit 1 higher order derivatives and applications

EM - 1

Unit – 1

Higher order derivatives and applications

Q 1. Compute limx→−2(3x2+5x−9)

Solution:

First, use property 2 to divide the limit into three separate limits. Then use property 1 to bring the constants out of the first two. This gives,

Limx→−2(3x2+5x−9) =limx→−2(3x2) + limx→−2(5x) −limx→−2(9)

=3(−2)2+5(−2) −(9)

= -7

Q 2. Find the limits of limx→3 [x(x+2)].

Solution: limx→3 [x(x+2)] = 3(3+2) = 3 x 5 = 15

Q 3: Differentiability Implies Continuity If f is differentiable at x0, then f is continuous at x0.

It turns out that equation (1) can be restated.

Definition. Derivatives in One Dimension - Alternate Definition

Solution: Let f : D ⊂ R → R and let x0 be an interior point of D.

Then f is differentiable at x0 if there is a number f ′ (x0) such that (2) lim x→x0 f(x) − f(x0) − (x − x0)f ′ (x0) |x − x0| = 0 and the number f ′ (x0) is called the derivative of f at x0.

We use this definition to create a definition of the derivative for functions of

Several variables. Definition. Derivatives in Higher Dimensions Let f : D ⊂ R n → R

And let P0 be an interior point of D. Then f is differentiable at P0 if there is a vector

f ′ (P0) such that

lim P→P0 f (P) − f (P0) − # » P0P · f ′ (P0) =0

|P − P0|

Q 4:Suppose there is a c∈Rc∈R and a function E:R→RE:R→R, s.t. f(x)=f(a)+c(x−a)+E(x)f(x)=f(a)+c(x−a)+E(x) near aa and limx→aE(x)x−a=0limx→aE(x)x−a=0.

Then:

Limx→af(x)−f(a)x−a−c=limx→af(x)−f(a)−c(x−a)x−a=0limx→af(x)−f(a)x−a−c=limx→af(x)−f(a)−c(x−a)x−a=0

So that means:

f′(x)f′(x) exists and f′(x)=cf′(x)=c. Of course writing f′(x)f′(x) instead of cc before you know this is indeed begging the question.

Q 5: For a function f: R→R f: R→R, suppose that there exist a real number AA such that for all xx in some neighbourhood of aa,

f(x)=f(a)+A(x−a) +E(x)f(x)=f(a)+A(x−a) + E(x)

And that

Limx → aE(x)(x−a) =0limx→aE(x)(x−a) =0

We say that ff is differentiable at aa. If ff is differentiable at aa we define f′(a)f′(a) by

f′(a)=Af′(a)=A

This definition is not circular and is useful since it can be generalized to the case where f: Rn→Rm f: Rn→Rm. In this case AA is a matrix instead of a real number (to be precise, AA is a linear operator