Differentiablity of sinx at pi/2

[R Power]

Is f(x)=sin x differentiable at x = pi/2. If yes why?

because what i think is at pi/2 f'(x) changes from 1 to -1 so at x=pi/2 it may not be differentiable (similar to case of |x|).

 

[Cyosis]

Why do you think f'(x) suddenly changes from 1 to -1 at pi/2?

 

[R Power]

zoom into the area where it changes sign ... there cerrtainly will be a region where sign will get changed ...it's obvious

 

[Cyosis]

There is an extremely major difference between gradually changing sign and a jump from 1 to -1 at pi/2 (as you claim). My question is why do you think it goes from 1 to -1 at x=pi/2?

 

[R Power]

i agree that it gradually changes sign (since the graph is curved and not straight as in |x|) . But in LHS and RHS of pi/2 slopes will be opposite ,not -1 and 1 but they will be opposite to each other (due to symmetry of curve) , so pi/2 is a point where slope changes from LHS to RHS. Thus it shouldn't be diferentiable at that point...

 

[Cyosis]

The signs will be opposite to each other yes, but they won't be -1 and 1 except for two points. Intuitively, if you can draw a unique tangent line to the curve at a point then the function has a derivative at that point. Can you draw a tangent line at the point x=Pi/2 for sin x? Can you draw one for the function |x| at x=0?

 

[R Power]

i understand what you are trying to say but i am still confused...

See, i'll tell you where I'm confused...
A function is differentiable at point xo if it the derivative is continuous as xo . Now derivative will be continuous at xo if it exists at neighbourhood points say x1 and x2 and is equal to it's value at xo. Right? Now consider point x1 (at Lhs of xo)... at Lhs of x1 say at x3 sllope will be same as that of x1 and xo(at RHS o x1) for derivative to be continuous at x1...now consider point x3 and repeat the same procedure and then x4 and u'll see that slope remains constant...

 

[HallsofIvy]

i understand what you are trying to say but i am still confused...

See, i'll tell you where I'm confused...
A function is differentiable at point xo if it the derivative is continuous as xo .

This is certainly NOT true. A function is differentiable at x0 if and only if its derivative exists- it is not necessary that the derivative be continuous there. A function is said to be "continuously differentible" at a point if the derivative is continuous there, as stronger condition than just being differentiable. And, in fact, sin(x) is "continuously differentiable" at \pi/2.

Now derivative will be continuous at xo if it exists at neighbourhood points say x1 and x2 and is equal to it's value at xo. Right?

No, not right! You seem to be saying that only constant functions are continuous! A function is continuous at x= a if the limit, as x goes to a is f(a). That, in turn, means that, given any \epsilon> 0, there exist a neighborhood in which |f(x)- f(a)|< \epsilon. It is certainly not necessary that any points in that neighborhood, other than a, have the same value as at a.

Now consider point x1 (at Lhs of xo)... at Lhs of x1 say at x3 sllope will be same as that of x1 and xo(at RHS o x1) for derivative to be continuous at x1...now consider point x3 and repeat the same procedure and then x4 and u'll see that slope remains constant...

Your are stating things about continuity and differentiability that are simply not true. You need to review both concepts.

The derivative of sin(x) is cos(x) and cos(\pi/2)= 0. That's how the derivative can change sign continuously at \pi/2.

 

[R Power]

This is certainly NOT true. A function is differentiable at x0 if and only if its derivative exists- it is not necessary that the derivative be continuous there. A function is said to be "continuously differentible" at a point if the derivative is continuous there, as stronger condition than just being differentiable. And, in fact, sin(x) is "continuously differentiable" at LaTeX Code: \\pi/2 .

can u give me example of a function whose derivative at particular point exists but the derivative is not continuous at that point?

No, not right! You seem to be saying that only constant functions are continuous! A function is continuous at x= a if the limit, as x goes to a is f(a). That, in turn, means that, given any LaTeX Code: \\epsilon> 0 , there exist a neighborhood in which LaTeX Code: |f(x)- f(a)|< \\epsilon . It is certainly not necessary that any points in that neighborhood, other than a, have the same value as at a.

When did i say only contant functions are continous. By stating points x1 and x2 i meant these points are infinitely close to xo at LHL and RHL, repectively. Lt x->a
means x comes very very closer to a. And that's what i meant!
Moreover, a function is continuous when LHL = F(a) = RHL.

 

[MichielM]

can u give me example of a function whose derivative at particular point exists but the derivative is not continuous at that point?

The example you gave in the first post is such a function: |x|.
That function is perfectly differentiable (it will be 1 or -1) throughout the domain, but it changes sign with 'infinite' slope at x=0.

 

[l'Hôpital]

can u give me example of a function whose derivative at particular point exists but the derivative is not continuous at that point?

<br /> f(x) = x^2 sin\frac{1}{x}<br />

Where f(0) = 0. Clearly, f is continuous at zero, as you can see by taking a limit. It is also differentiable at zero, as you can see by doing the definition.
<br /> \lim_{h \rightarrow 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{f(h) - 0}{h}<br />
<br /> \lim_{h \rightarrow 0} \frac{h^2 sin\frac{1}{h}}{h} = \lim_{h \rightarrow 0} h sin \frac{1}{h} = 0<br />

So f&#039;(0) = 0

But...
<br /> f&#039;(x) = 2x sin \frac{1}{x} - cos\frac{1}{x}<br />
Which is obviously discontinuous at x = 0.

 

[HallsofIvy]

The example you gave in the first post is such a function: |x|.
That function is perfectly differentiable (it will be 1 or -1) throughout the domain, but it changes sign with 'infinite' slope at x=0.

No. |x| is NOT differentiable at x= 0.

The function L'Hopital gives, above, x^2 sin(\frac{1}{x}), is differentiable for all x but the derivative is not continuous at x= 0.

 

[HallsofIvy]

can u give me example of a function whose derivative at particular point exists but the derivative is not continuous at that point?

L'Hopital gave an example above: x^2 sin(\frac{1}{x}) is continuous for all x butt the derivative is not continuous at x= 0.

When did i say only contant functions are continous.

You said "Now derivative will be continuous at xo if it exists at neighbourhood points say x1 and x2 and is equal to it's value at xo." If, for any neighborhood, there exist points x1 and x2 such that f(x1)= f(x2)= f(x0), then f must be constant near x0.

By stating points x1 and x2 i meant these points are infinitely close to xo at LHL and RHL, repectively. Lt x->a
means x comes very very closer to a. And that's what i meant!
Moreover, a function is continuous when LHL = F(a) = RHL.

There are no such thing as "points infinitely close to x0". You have a very strange notion of what a limit is. "Let x-> a" does NOT mean "x comes very very close to a". You seem to be trying to talk about infinitesmals, a very deep and abstract way to talk about limits, but I suspect that you are not clear on what they are.

 

[Char. Limit]

Was it necessary to post the same content three times?

sin(x) is differentiable for all x because cos(x) exists for all x. Simple.

 

[Anonymous217]

^ It was just a lag issue. That may happen when you post a reply, the page starts to load, you input a few more words, you click it again, and etc.

zoom into the area where it changes sign ... there cerrtainly will be a region where sign will get changed ...it's obvious

If it's obvious that it's changing sign gradually can't you see that the limits of both the right hand and the left hand of f'(x) would reach 0? That is, the slope approaches 0 as x reaches pi/2.

 

[R Power]

OK now i understand i think!
In function sinx slope gradually changes so at pi/2 we have only one tangent whose slope is 0.
Whereas in functions like |x| at x=0 infinite tangents in between -1 and 1 are possible .Hence derivative is not defined at x=0.
Am i right now?