Is XSIN 1 x is differentiable?

Is XSIN 1 x is differentiable?

Prove it is a differentiable function. x→0(xsin 1x ). x→0(xsin 1x )=0. (8) Therefore f(x) is differentiable at x=0 and f′(0) = 0.

Where is sin 1 x differentiable?

The function f(x)=sin(1/x) is differentiable on it’s domain. The derivative is f′(x)=cos(1/x)ddx1x=−1x2cos(1x).

How do you know if a function is differentiable at 0?

0 if x ≤ 0. f(h) h . f(h) h = 0. Since the left and right limits exist and are equal, the limit also exists, and f is differentiable at 0 with f (0) = 0.

Does the function f/x XSIN 1 x have a removable discontinuity at x 0?

The function f(x)=xsin(1/x) is not 0 at x=0 as it is not even defined there. But it does have a removable discontinuity there, i.e. limx→0xsin(1/x)=0. You can easily prove this using Squeeze Theorem, comparing f(x) to |x| because |sin(1/x)|≤1.

Is the function XSIN 1 x continuous?

The function sin(1/x). The function is odd, f(−x) = −f(x), its graph is symmetric with respect to the origin (0, 0) . As a composition of 1/x and sinx, f(x) is continuous at each point of its domain.

Does XSIN 1 x have a removable discontinuity?

So it isn’t a removable discontinuity. One might call this an essential discontinuity.

Is sine function always differentiable?

Theorem The function sin x is differentiable everywhere, and its derivative is cos x. Indeed, you can see that when sin x has zero slope cosine has value 0, and so on.

What is the domain of differentiability of sin inverse X?

The domain of sin−1 is [−1,1] and its range is [−π2,π2]. We can see from the graph that sin−1 is an odd function, that is, sin−1(−x)=−sin−1x.

How do you know if a function is differentiable?

A function is said to be differentiable if the derivative of the function exists at all points in its domain. Particularly, if a function f(x) is differentiable at x = a, then f′(a) exists in the domain.

How do you check a function is differentiable or not?

A function is formally considered differentiable if its derivative exists at each point in its domain, but what does this mean? It means that a function is differentiable everywhere its derivative is defined. So, as long as you can evaluate the derivative at every point on the curve, the function is differentiable.

Does this function have a discontinuity at x 0?

The function 1/x is continuous on (0, ∞) and on (−∞, 0), i.e., for x > 0 and for x < 0, in other words, at every point in its domain. However, it is not a continuous function since its domain is not an interval. It has a single point of discontinuity, namely x = 0, and it has an infinite discontinuity there.

Is X 2sin 1 x uniformly continuous?

The function g is continuous. Therefore it is uniformly continuous on any compact set. g′(x)=2xsin(1x)−cos(1x).