Proving Differentiability and Continuity of f'(x)

In summary, the given function is differentiable everywhere and its derivative is continuous. The derivative is 0 when x is less than or equal to 0, and when x is greater than 0, the derivative can be found using the product rule and chain rule. By showing that the limit of the derivative as x approaches 0 from the right is equal to the derivative at 0, we can conclude that the derivative is also continuous at 0.
  • #1
Edellaine
11
0

Homework Statement


show

[tex]f(x)=\left\{e^{\frac{-1}{x}} \\\ x>0[/tex]
[tex]f(x)=\left\{0 \\\ x\leq 0[/tex]
is differentiable everywhere, and show its derivative is continuous


Homework Equations


Product Rule and Chain Rule for derivatives. Definition of a derivative
[tex]f^{'}(a)=\frac{f(x)-f(a)}{x-a}[/tex]


The Attempt at a Solution


Showing that f(x) is differentiable when x > 0 is easy. I use product rule and chain rule to define functions that when composed make f(x). Then for the derivative of f(x) when x>0, I show f'(x) is continuous by finding f''(x) which implies f'(x) is continuous.

I run into problems when x[tex]\leq[/tex]0.

I say on x[tex]\leq[/tex]0, [tex]f^{'}(o)=\frac{f(x)-f(o)}{x-o}=\frac{0}{x}=0[/tex] as x[tex]\rightarrow0[/tex] so f'(0) exists.
So now f'(x)=0 for all x < 0, and f'(x)[tex]\rightarrow[/tex]0 as x[tex]\rightarrow[/tex]0.

I run into my problem when trying to show that the derivative here is continuous. I'm not sure where to go from here. I'm really blanking and haven't really even made an attempt at a solution. My only idea is to maybe use right-hand and left-hand limits to show that the limit of f'(x) at 0 exists, so therefore it is also continuous at 0. I really feel I'm reaching here, and I'd appreciate some guidance.

Thanks.
 
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  • #2


You're right in that you need to show that the limit as x-->0 of f'(x) is f'(0) in order to show that the derivative is continuous. In order to do this you need to find the derivative...clearly it is 0 for x[tex]\leq[/tex]0...what about for x>0?

Differentiate, then show that the derivative approaches 0 as x-->0+. It should be clear that the derivative is continuous for x[tex]\neq[/tex]0 and left-continuous at x=0, so all you need to do is show right continuity at 0.
 

1. How do I prove that f'(x) is differentiable at a certain point?

To prove that f'(x) is differentiable at a point x=a, you need to show that the following limit exists and is finite:lim(h→0) [(f(a+h) - f(a))/h]

2. What is the difference between differentiability and continuity?

Continuity refers to a function being continuous at a point, which means that the limit of the function as x approaches that point exists and is equal to the function value at that point. Differentiability refers to a function being differentiable at a point, which means that the limit of the function's derivative as x approaches that point exists and is finite.

3. Can a function be differentiable at a point but not continuous?

No, a function cannot be differentiable at a point if it is not continuous at that point. If a function is differentiable at a point, it must also be continuous at that point.

4. What are the different ways to prove differentiability and continuity of f'(x)?

There are a few different methods for proving differentiability and continuity of f'(x). Some common approaches include using the definition of a derivative, using the differentiability criteria for different types of functions (such as polynomials or trigonometric functions), and using the continuity and differentiability rules for composite functions.

5. How do I prove differentiability and continuity of f'(x) for a piecewise function?

To prove differentiability and continuity of f'(x) for a piecewise function, you will need to consider each piece of the function separately and show that it is differentiable and continuous at the points where the pieces are connected. You may need to use different methods depending on the types of functions involved in each piece.

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