Y= { x if x<_0 , 2x if x>_ 0 } find if y is differentiable at 0 .Then, if y is differentiable find the derivative at x=0.
Y’ = { 1 if x < 0 ,2 if x >0 } How do you continue from here?
Since
y' = 1 for x <= 0
y' = 2 for x>0
y' is not continuous at x=9. So, y is not differentiable there.
Whenever you have a cusp or point on a graph, the function is not differentiable there.
Better reread the definition of differentiable.
It is looking for y to be differentiable at 0. I think I understand because it is not continuous at x=0.
To determine if y is differentiable at x = 0, we need to check if the left-hand derivative at 0 is equal to the right-hand derivative at 0, and if they both exist.
First, let's find the left-hand derivative. Since y(x) is defined as x for x < 0, the left-hand derivative can be found by evaluating the derivative of x at x = 0. Taking the derivative of x with respect to x gives us 1.
Next, let's find the right-hand derivative. We have y(x) defined as 2x for x > 0. Taking the derivative of 2x with respect to x also gives us 2.
Now, we check if the left-hand derivative at 0 (which is 1) is equal to the right-hand derivative at 0 (which is 2). Since they are not equal, y is not differentiable at x = 0.
Therefore, y is not differentiable at x = 0, and we cannot find the derivative at x = 0 as it does not exist.