# Question Regarding derivatives

• Mar 11th 2010, 09:54 AM
Solid8Snake
Question Regarding derivatives
Greeting I have a question regarding continuity and derivatives can someone help me with this question if possible. The question I am referring to is 6).
• Mar 12th 2010, 12:02 AM
Sudharaka
Dear Solid8Snake,

If x<-1, $\displaystyle \frac{dy}{dx}=-2$

If -1<x<1, $\displaystyle \frac{dy}{dx}=2x$

If x>1, $\displaystyle \frac{dy}{dx}=1$

Therefore f is differentiable throught the real number system.

Can you do it from here?
• Mar 12th 2010, 03:41 AM
HallsofIvy
Quote:

Originally Posted by Sudharaka
Dear Solid8Snake,

If x<-1, $\displaystyle \frac{dy}{dx}=-2$

If -1<x<1, $\displaystyle \frac{dy}{dx}=2x$

If x>1, $\displaystyle \frac{dy}{dx}=1$

Therefore f is differentiable throught the real number system.

This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\displaystyle \lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

From below that is $\displaystyle \lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\to 0}\frac{(1+h)^2- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{2h+ h^2}{h}= 2. From above that is \displaystyle \lim_{h\to 0^+}\frac{f(1+h)- f(1)}{h}= \lim_{h\to 0}\frac{1+h- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{h}{h}= 1$.

Since those two one sided limits are not the same the limit itself and so the derivative does not exist at x= 1.

While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so if the limits of the derivative function, from either side, exist, they must be the same. Here, the limit of 2x, as x goes to 1 is 2 and the limit of 1, as x goes to 1, is 1. They exist but are not the same so the function is not differentiable at x= 1.

There will be a "corner" at (1, 1) with the curve coming in from the left at angle $\displaystyle tan(\theta)= 2$ and from the right at angle $\displaystyle tan(\theta)= 1$.

Quote:

Can you do it from here?
• Mar 12th 2010, 03:43 AM
HallsofIvy
Quote:

Originally Posted by Sudharaka
Dear Solid8Snake,

If x<-1, $\displaystyle \frac{dy}{dx}=-2$

If -1<x<1, $\displaystyle \frac{dy}{dx}=2x$

If x>1, $\displaystyle \frac{dy}{dx}=1$

Therefore f is differentiable throught the real number system.

This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\displaystyle \lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

From below that is $\displaystyle \lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\to 0}\frac{(1+h)^2- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{2h+ h^2}{h}= 2. From above that is \displaystyle \lim_{h\to 0^+}\frac{f(1+h)- f(1)}{h}= \lim_{h\to 0}\frac{1+h- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{h}{h}= 1$.

Since those two one sided limits are not the same the limit itself and so the derivative does not exist at x= 1.

While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so if the limits of the derivative function, from either side, exist, they must be the same. Here, the limit of 2x, as x goes to 1 is 2 and the limit of 1, as x goes to 1, is 1. They exist but are not the same so the function is not differentiable at x= 1.

Notice that at x= -2, the limits of the two derivative functions are $\displaystyle \lim_{x\to -1} -2= -2$ and $\displaystyle \lim_{x\to -1} 2x= -2$ so the function is differentiable at x= -1.

There will be a "corner" in the graph at (1, 1) with the curve coming in from the left at angle $\displaystyle tan(\theta)= 2$ and from the right at angle $\displaystyle tan(\theta)= 1$.

Quote:

Can you do it from here?
• Mar 12th 2010, 06:05 PM
Sudharaka
Quote:

Originally Posted by HallsofIvy
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\displaystyle \lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

From below that is $\displaystyle \lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\to 0}\frac{(1+h)^2- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{2h+ h^2}{h}= 2. From above that is \displaystyle \lim_{h\to 0^+}\frac{f(1+h)- f(1)}{h}= \lim_{h\to 0}\frac{1+h- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{h}{h}= 1$.

Since those two one sided limits are not the same the limit itself and so the derivative does not exist at x= 1.

While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so if the limits of the derivative function, from either side, exist, they must be the same. Here, the limit of 2x, as x goes to 1 is 2 and the limit of 1, as x goes to 1, is 1. They exist but are not the same so the function is not differentiable at x= 1.

Notice that at x= -2, the limits of the two derivative functions are $\displaystyle \lim_{x\to -1} -2= -2$ and $\displaystyle \lim_{x\to -1} 2x= -2$ so the function is differentiable at x= -1.

There will be a "corner" in the graph at (1, 1) with the curve coming in from the left at angle $\displaystyle tan(\theta)= 2$ and from the right at angle $\displaystyle tan(\theta)= 1$.

Dear HallsofIvy,

Thank you very much for showing my mistake. This is what I always expect from MHF, to know things I don't know and to further improve the knowledge of what I already know. Thanks a lot. (I suppose you are not angry with me for posting an incorrect answer!!)

Goodbye.