Mixing
Contradiction between first derivative formal definition and derivative rules?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
When I try to find the derivative of $f(x) = sqrt[3]x sin(x)$ at $x=0$, using the formal definition of first derivative, I get this:
$
f'(0) = largelimlimits_x to 0 Largefracsqrt[3]x sin(x)-0x-0$, which gives zero.
However, when I use derivative rules i get that:
$
f'(x) = large sin(x) frac13sqrt[3]x^2+cos(x)sqrt[3]x$
and thus $f'(0)$ doesn't exist, why does this happen? what's the reason behind it?
limits derivatives
edited 1 hour ago
Key Flex
6,36321028
asked 1 hour ago
Just_Cause
183
add a comment |Â
up vote
3
down vote
favorite
When I try to find the derivative of $f(x) = sqrt[3]x sin(x)$ at $x=0$, using the formal definition of first derivative, I get this:
$
f'(0) = largelimlimits_x to 0 Largefracsqrt[3]x sin(x)-0x-0$, which gives zero.
However, when I use derivative rules i get that:
$
f'(x) = large sin(x) frac13sqrt[3]x^2+cos(x)sqrt[3]x$
and thus $f'(0)$ doesn't exist, why does this happen? what's the reason behind it?
limits derivatives
edited 1 hour ago
Key Flex
6,36321028
asked 1 hour ago
Just_Cause
183
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
When I try to find the derivative of $f(x) = sqrt[3]x sin(x)$ at $x=0$, using the formal definition of first derivative, I get this:
$
f'(0) = largelimlimits_x to 0 Largefracsqrt[3]x sin(x)-0x-0$, which gives zero.
However, when I use derivative rules i get that:
$
f'(x) = large sin(x) frac13sqrt[3]x^2+cos(x)sqrt[3]x$
and thus $f'(0)$ doesn't exist, why does this happen? what's the reason behind it?
limits derivatives
edited 1 hour ago
Key Flex
6,36321028
asked 1 hour ago
Just_Cause
183
When I try to find the derivative of $f(x) = sqrt[3]x sin(x)$ at $x=0$, using the formal definition of first derivative, I get this:
$
f'(0) = largelimlimits_x to 0 Largefracsqrt[3]x sin(x)-0x-0$, which gives zero.
However, when I use derivative rules i get that:
$
f'(x) = large sin(x) frac13sqrt[3]x^2+cos(x)sqrt[3]x$
and thus $f'(0)$ doesn't exist, why does this happen? what's the reason behind it?
limits derivatives
limits derivatives
edited 1 hour ago
Key Flex
6,36321028
asked 1 hour ago
Just_Cause
183
edited 1 hour ago
Key Flex
6,36321028
asked 1 hour ago
Just_Cause
183
edited 1 hour ago
Key Flex
6,36321028
edited 1 hour ago
Key Flex
6,36321028
edited 1 hour ago
Key Flex
6,36321028
6,36321028
asked 1 hour ago
Just_Cause
183
asked 1 hour ago
Just_Cause
183
asked 1 hour ago
Just_Cause
183
183
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $sqrt[3]x$ is not differentiable at $x=0$.
answered 1 hour ago
ajotatxe
51.5k23286
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It makes sense now.
– Just_Cause
51 mins ago
add a comment |Â
up vote
1
down vote
Your computations of derivatives are correct. However, be careful this function is not differentiable at zero (one finds problems with 0/0 etc).
Never lose faith in the formal definitions :)
answered 1 hour ago
analytic
320111
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $sqrt[3]x$ is not differentiable at $x=0$.
answered 1 hour ago
ajotatxe
51.5k23286
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It makes sense now.
– Just_Cause
51 mins ago
add a comment |Â
up vote
4
down vote
accepted
The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $sqrt[3]x$ is not differentiable at $x=0$.
answered 1 hour ago
ajotatxe
51.5k23286
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It makes sense now.
– Just_Cause
51 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $sqrt[3]x$ is not differentiable at $x=0$.
answered 1 hour ago
ajotatxe
51.5k23286
The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $sqrt[3]x$ is not differentiable at $x=0$.
answered 1 hour ago
ajotatxe
51.5k23286
answered 1 hour ago
ajotatxe
51.5k23286
answered 1 hour ago
ajotatxe
51.5k23286
answered 1 hour ago
ajotatxe
51.5k23286
51.5k23286
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It makes sense now.
– Just_Cause
51 mins ago
add a comment |Â
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It makes sense now.
– Just_Cause
51 mins ago
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition.
– Just_Cause
54 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It is possible that $g$ is not differentiable and $fg$ is.
– ajotatxe
52 mins ago
It makes sense now.
– Just_Cause
51 mins ago
It makes sense now.
– Just_Cause
51 mins ago
add a comment |Â
up vote
1
down vote
Your computations of derivatives are correct. However, be careful this function is not differentiable at zero (one finds problems with 0/0 etc).
Never lose faith in the formal definitions :)
answered 1 hour ago
analytic
320111
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
add a comment |Â
up vote
1
down vote
Your computations of derivatives are correct. However, be careful this function is not differentiable at zero (one finds problems with 0/0 etc).
Never lose faith in the formal definitions :)
answered 1 hour ago
analytic
320111
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your computations of derivatives are correct. However, be careful this function is not differentiable at zero (one finds problems with 0/0 etc).
Never lose faith in the formal definitions :)
answered 1 hour ago
analytic
320111
Your computations of derivatives are correct. However, be careful this function is not differentiable at zero (one finds problems with 0/0 etc).
Never lose faith in the formal definitions :)
answered 1 hour ago
analytic
320111
edited 29 mins ago
answered 1 hour ago
analytic
320111
answered 1 hour ago
analytic
320111
answered 1 hour ago
analytic
320111
320111
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
add a comment |Â
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
You mean the cubic root of x that's not differentiable at x = 0 or the whole function?
– Just_Cause
54 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
In this case the cubic root of $x$ is not differentiable at zero and that causes the whole function to be not differentiable at zero. The reason is because the derivative of $x^1/3$ is $(1/3)x^-2/3$ and this would mean trying to divide by zero when $x=0$ - which is impossible!
– analytic
50 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
Aha exactly, so it turns out that derivative rules aren't flawless like formal definition.
– Just_Cause
48 mins ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2990086%2fcontradiction-between-first-derivative-formal-definition-and-derivative-rules%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
