# 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).