Calc 1 - derivative of absolute value

[hayesk85]

Homework Statement

Question is: how can you tell if there are any places you can't take the derivative of an equation that has an absolute value (using logic, not just graphing it)

example equations

1. [tex]\left|[/tex]x-5[tex]\right|[/tex]

2. [tex]\left|[/tex] x³+4x²+9x+17 [tex]\right|[/tex]
x²+1

3. [tex]\left|[/tex][tex]\sqrt[3]{x}[/tex][tex]\right|[/tex]

The Attempt at a Solution

1. I realize there is a corner at x=5, but the limit still exists there and the limits match

 

[danago]

Homework Statement

Question is: how can you tell if there are any places you can't take the derivative of an equation that has an absolute value (using logic, not just graphing it)

example equations

1. [tex]\left| x-5 \right|[/tex]

1. I realize there is a corner at x=5, but the limit still exists there and the limits match

Are the limits really equal from both sides? Remember that as x-->5, the value of x-5 could be positive or negative, depending on which side of 5 x is approaching from. So if x-->5+, then x is getting close to 5, but is SLIGHTLY MORE, hence the quantity x-5 will be positive. If we had x-->5-, then x is will always be SLIGHTLY LESS than 5, making x-5 negative. For this reason, the left and right hand derivatives are not infact equal.

 

[HallsofIvy]

If f(x)= |x-5| the
[tex]\frac{df}{dx}(5)= \lim_{h\rightarrow 0}\frac{f(5+h)- f(5)}{h}= \lim_{h\rightarrow 0}\frac{|5+h- 5|}{h}= \lim_{h\rightarrow 0}\frac{|h|}{h}[/tex]

If h< 0, that fraction is -1, if h> 0, that fraction is 1. Yes, the two limits exist. No, they are not the same.

It is impossible to tell what you meant by the others. Do not use "tex" or "itex" on individual parts- include the entire formula. Here, I think, you have "\left|" in one "tex" formula and "\right|" in another. They have to be balanced in the same "tex" formula.
I still couldn't tell what you meant since you seem to be using an underline, "\U" to get fractions. Use \frac{}{} instead.

More generally, if f(x)= |g(x)|, by the chain rule, taking u= g(x),
[tex]\frac{df}{dx}= \frac{d|u|}{du}\frac{du}{dx}[/tex]
The first derivative, d|u|/du, does not exist wherever u= g(x)= 0 and, so, neither does df/dx.