- #1

Hummingbird25

- 86

- 0

**An analytical approach to the differential equation(urgent)**

## Homework Statement

(1) Given the function [tex]f: t \mapsto |t|[/tex], show that the function f is a differentiable function where [tex]t \neq 0[/tex] and write f's the differential coffiecient.

2) Given the function [tex]g: t \mapsto \sqrt{t^2}[/tex], show that g is differentiable if [tex]t \neq 0[/tex], and write the differentiable coeffient of g.

## Homework Equations

Also my textbook asks why can't you use the chainrule to find the differential coefficient at t = 0. Isn't that because the limit doesn't exist?

## The Attempt at a Solution

My Solution part(1):

*According to the definition of my textbook for a function to be a differentiable function the following must be meet:*

(1) Let [tex]A \subseteq \mathbb{R}[/tex], let [tex]f: A \rightarrow \mathbb{R}[/tex] be a real function defined on A, and let 'a' be a point in A.

Then f is a differentiable function at a, if t is an inner point of A and the differential coeffient is

[tex]\frac{f(t)-f(a)}{t-a}[/tex] has a finit limit where t tends towards a.

Therefore [tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a}[/tex]

(1) Let [tex]A \subseteq \mathbb{R}[/tex], let [tex]f: A \rightarrow \mathbb{R}[/tex] be a real function defined on A, and let 'a' be a point in A.

Then f is a differentiable function at a, if t is an inner point of A and the differential coeffient is

[tex]\frac{f(t)-f(a)}{t-a}[/tex] has a finit limit where t tends towards a.

Therefore [tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a}[/tex]

Assuming that t = 0, then it an inner point which implies that exist an r > 0 [tex]\Rightarrow B_{r}(t = 0) \in A[/tex]. (Refer to the definition of the inner point). But according of the limit, this doesn't exist then t = 0.

So therefore

[tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a} = \mathop {\lim }\limits_{t \to a}\frac{|t|-|a|}{t-a}[/tex] is the differential coeffient is required which is only valied at [tex]t \neq 0.[/tex]

Assuming that t = 0, then it an inner point which implies that exist an r > 0 [tex]\Rightarrow B_{r}(t = 0) \in A[/tex]. (Refer to the definition of the inner point). But according of the limit, this doesn't exist then t = 0.

My solution(2)

2) Given the function [tex]g: t \mapsto \sqrt{t^2}[/tex], show that g is differentiable if [tex]t \neq 0[/tex], and write the differentiable coeffient of g.

As in (1) x = 0 is assumed not be an inner point, and therefore its not possible to find the differentiable coeffient at that point?

Then the differentiable coeffient is

[tex]g'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a} = \mathop {\lim }\limits_{t \to a}\frac{|t|-|a|}{t-a}[/tex]

Since [tex]\sqrt{t^2} = |t|[/tex] according to My TI-92 calculator.

Isn't this verified by

[tex]g'(t) = \frac{2t}{2 \cdot \sqrt{t^2}}[/tex] where [tex]g(t) = \int (g'(t)) dt = |t| [/tex]

Cheers Hummingbird

Last edited: