Your function separates into two linear functions,
if x > 0 we get y = 2x -1 for a slope of 2
if x < 0 we get y = 4x -1 for a slope of 4
of course at (0,-1) we have a problem.
so we can find the derivative for all values of x, except x = 0
for y = x^(2/5)
dy/dx = (2/5)x^(-3/5)
= 2/[5x^(3/5)]
which is what you have.
What do you mean "now what?"
What do you want to do with it?
determine if differentiable everywhere:
3x-abs(x)-1=y
i know you're supposed to find the derivative.... -- then my mind draws a blank? hahah please help me out
also
y=x ^ 2/5
you find the derivative and it is 2/(5x^(3/5)) now what?
thanks so much
3 answers
determine if differentiable everywhere:`
"everywhere" means the ℝ domain, or [-∞,∞].
If you can find a point where the function is not differentiable, then the function is not differentiable everywhere.
For a function to be "differentiable" in an interval, it needs to be continuous within the interval. This is a necessary condition.
Is the given function continuous? If not, it is not diffentiable where discontinuity occurs.
Another necessary condition for differentiability is that for every point in the interval, either
1. the derivative exists, or
2. both the left-hand and right-hand derivatives exist, and they are equal.
For
y=3x-abs(x)-1, see
http://img23.imageshack.us/img23/4836/1254714362andre1.png
y' = 3-x/abs(x) which is a step function at x=0, which means that at x=0, the derivatives from the left and right are 4 and 2 respectively. Make your conclusion whether the function is differentiable "everywhere", and if not, define the interval in which it is.
For the other function,
y=x^(2/5)
make a table of values of the function for x=-3 to +3 in steps of 1 and confirm that
1. it is continuous between -3 and +3
2. it is an even function
Also calculate y' at 0+ and 0-.
Conclude if the function is differentiable "everywhere".
See:
http://img23.imageshack.us/img23/4836/1254714362andre.png
"everywhere" means the ℝ domain, or [-∞,∞].
If you can find a point where the function is not differentiable, then the function is not differentiable everywhere.
For a function to be "differentiable" in an interval, it needs to be continuous within the interval. This is a necessary condition.
Is the given function continuous? If not, it is not diffentiable where discontinuity occurs.
Another necessary condition for differentiability is that for every point in the interval, either
1. the derivative exists, or
2. both the left-hand and right-hand derivatives exist, and they are equal.
For
y=3x-abs(x)-1, see
http://img23.imageshack.us/img23/4836/1254714362andre1.png
y' = 3-x/abs(x) which is a step function at x=0, which means that at x=0, the derivatives from the left and right are 4 and 2 respectively. Make your conclusion whether the function is differentiable "everywhere", and if not, define the interval in which it is.
For the other function,
y=x^(2/5)
make a table of values of the function for x=-3 to +3 in steps of 1 and confirm that
1. it is continuous between -3 and +3
2. it is an even function
Also calculate y' at 0+ and 0-.
Conclude if the function is differentiable "everywhere".
See:
http://img23.imageshack.us/img23/4836/1254714362andre.png
The first link (for y = 3x-abs(x)-1 ) should read:
(Broken Link Removed)
(Broken Link Removed)
1 answer