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Why we can't differentiate a polynomial equation as many times as we wish?
Proof for exact differential equations shortcut?Implicit differentiation. If $sin y=2sin x$, show $(fracdydx)^2=1 + 3sec^2y$Sketch parametric curveWhy can't one implicitly differentiate these two relations?Simple system of two nonhomogeneous ordinary differential equations solved by elimination. (Ex 3.1-2)How to solve this implicit differentiation problem concerning arcsin?Solving $sin x = x^3-2x^2+1$ using Newton's MethodWhen is it allowed to do operations like 'differentiating both sides', 'integrating both sides'?A weird differentiation question.Why can't I sub $a=0$ into this expression?
$begingroup$
Suppose we had the equation below and we are going to differentiate it both sides:
beginalign
&2x^2-x=1\
&4x-1=0\
&4=0
endalign
This problem doesn't seems to happens with other equation like $ln x =1$ or $sin x = 0$, we can keep differentiating these two without getting "$4=0$", for example. This why I asked about polynomials.
PS: I'm not trying to solve any of these equations by differentiating then. But differentiation or integration helps and solving equations?
I remember that sometimes to solve trigonometry equtions like $sin x = cos x$ we had to square both side so we could use the identity $sin^2x + cos^2x =1$. Even thought squaring appears to make it worse because we have a new root.
real-analysis calculus derivatives
asked 1 hour ago
PintecoPinteco
796313
$endgroup$
add a comment |
$begingroup$
Suppose we had the equation below and we are going to differentiate it both sides:
beginalign
&2x^2-x=1\
&4x-1=0\
&4=0
endalign
This problem doesn't seems to happens with other equation like $ln x =1$ or $sin x = 0$, we can keep differentiating these two without getting "$4=0$", for example. This why I asked about polynomials.
PS: I'm not trying to solve any of these equations by differentiating then. But differentiation or integration helps and solving equations?
I remember that sometimes to solve trigonometry equtions like $sin x = cos x$ we had to square both side so we could use the identity $sin^2x + cos^2x =1$. Even thought squaring appears to make it worse because we have a new root.
real-analysis calculus derivatives
asked 1 hour ago
PintecoPinteco
796313
$endgroup$
12
$begingroup$
The function $2x^2-2$ is not the same as the function $1$ so it makes no sense to differentiate both sides of that equation the way you have.
$endgroup$
– lulu
1 hour ago
1
$begingroup$
$x=1$ and $x=0$ don't work, either. Differentiating both sides gives you $1=0.$ @Pinteco In general, an "equation" is one where we are trying to solve for individual values, but differentiation requires values in an area around the value for $x,$ so in general, if you are trying to solve $f(x)=g(x),$ you cannot differential both sides and get an equation. If, however, for every $x$ in an interval, you have $f(x)=g(x)$, then you can differentiate both sides and still get an equation, potentially more solutions, but containing the solutions in that interval.
$endgroup$
– Thomas Andrews
1 hour ago
$begingroup$
You can indeed take derivatives, like any other function, as much as you want, on both sides of an equality between objects for which the operation of taking derivatives is defined. The objects being equated in the first equality are not functions, but constant numbers. You could, if you want, tread them as constant functions of some other variable $y$ and then take derivative with respect to $y$. This would give you the true equation $0=0$.
$endgroup$
– user647486
1 hour ago
$begingroup$
Instead you computed as if taking derivative with respect to $x$. Derivative with respect to $x$ is defined for some functions of $x$, . But that is not an equality between functions of $x$. Equality of functions, by definition, is an equation that is satisfied for all values of $x$.
$endgroup$
– user647486
1 hour ago
add a comment |
$begingroup$
Suppose we had the equation below and we are going to differentiate it both sides:
beginalign
&2x^2-x=1\
&4x-1=0\
&4=0
endalign
This problem doesn't seems to happens with other equation like $ln x =1$ or $sin x = 0$, we can keep differentiating these two without getting "$4=0$", for example. This why I asked about polynomials.
PS: I'm not trying to solve any of these equations by differentiating then. But differentiation or integration helps and solving equations?
I remember that sometimes to solve trigonometry equtions like $sin x = cos x$ we had to square both side so we could use the identity $sin^2x + cos^2x =1$. Even thought squaring appears to make it worse because we have a new root.
real-analysis calculus derivatives
asked 1 hour ago
PintecoPinteco
796313
$endgroup$
Suppose we had the equation below and we are going to differentiate it both sides:
beginalign
&2x^2-x=1\
&4x-1=0\
&4=0
endalign
This problem doesn't seems to happens with other equation like $ln x =1$ or $sin x = 0$, we can keep differentiating these two without getting "$4=0$", for example. This why I asked about polynomials.
PS: I'm not trying to solve any of these equations by differentiating then. But differentiation or integration helps and solving equations?
I remember that sometimes to solve trigonometry equtions like $sin x = cos x$ we had to square both side so we could use the identity $sin^2x + cos^2x =1$. Even thought squaring appears to make it worse because we have a new root.
real-analysis calculus derivatives
real-analysis calculus derivatives
asked 1 hour ago
PintecoPinteco
796313
asked 1 hour ago
PintecoPinteco
796313
asked 1 hour ago
PintecoPinteco
796313
asked 1 hour ago
PintecoPinteco
796313
asked 1 hour ago
PintecoPinteco
796313
796313
12
$begingroup$
The function $2x^2-2$ is not the same as the function $1$ so it makes no sense to differentiate both sides of that equation the way you have.
$endgroup$
– lulu
1 hour ago
1
$begingroup$
$x=1$ and $x=0$ don't work, either. Differentiating both sides gives you $1=0.$ @Pinteco In general, an "equation" is one where we are trying to solve for individual values, but differentiation requires values in an area around the value for $x,$ so in general, if you are trying to solve $f(x)=g(x),$ you cannot differential both sides and get an equation. If, however, for every $x$ in an interval, you have $f(x)=g(x)$, then you can differentiate both sides and still get an equation, potentially more solutions, but containing the solutions in that interval.
$endgroup$
– Thomas Andrews
1 hour ago
$begingroup$
You can indeed take derivatives, like any other function, as much as you want, on both sides of an equality between objects for which the operation of taking derivatives is defined. The objects being equated in the first equality are not functions, but constant numbers. You could, if you want, tread them as constant functions of some other variable $y$ and then take derivative with respect to $y$. This would give you the true equation $0=0$.
$endgroup$
– user647486
1 hour ago
$begingroup$
Instead you computed as if taking derivative with respect to $x$. Derivative with respect to $x$ is defined for some functions of $x$, . But that is not an equality between functions of $x$. Equality of functions, by definition, is an equation that is satisfied for all values of $x$.
$endgroup$
– user647486
1 hour ago
add a comment |
12
$begingroup$
The function $2x^2-2$ is not the same as the function $1$ so it makes no sense to differentiate both sides of that equation the way you have.
$endgroup$
– lulu
1 hour ago
1
$begingroup$
$x=1$ and $x=0$ don't work, either. Differentiating both sides gives you $1=0.$ @Pinteco In general, an "equation" is one where we are trying to solve for individual values, but differentiation requires values in an area around the value for $x,$ so in general, if you are trying to solve $f(x)=g(x),$ you cannot differential both sides and get an equation. If, however, for every $x$ in an interval, you have $f(x)=g(x)$, then you can differentiate both sides and still get an equation, potentially more solutions, but containing the solutions in that interval.
$endgroup$
– Thomas Andrews
1 hour ago
$begingroup$
You can indeed take derivatives, like any other function, as much as you want, on both sides of an equality between objects for which the operation of taking derivatives is defined. The objects being equated in the first equality are not functions, but constant numbers. You could, if you want, tread them as constant functions of some other variable $y$ and then take derivative with respect to $y$. This would give you the true equation $0=0$.
$endgroup$
– user647486
1 hour ago
$begingroup$
Instead you computed as if taking derivative with respect to $x$. Derivative with respect to $x$ is defined for some functions of $x$, . But that is not an equality between functions of $x$. Equality of functions, by definition, is an equation that is satisfied for all values of $x$.
$endgroup$
– user647486
1 hour ago
12
12
$begingroup$
The function $2x^2-2$ is not the same as the function $1$ so it makes no sense to differentiate both sides of that equation the way you have.
$endgroup$
– lulu
1 hour ago
$begingroup$
The function $2x^2-2$ is not the same as the function $1$ so it makes no sense to differentiate both sides of that equation the way you have.
$endgroup$
– lulu
1 hour ago
1
1
$begingroup$
$x=1$ and $x=0$ don't work, either. Differentiating both sides gives you $1=0.$ @Pinteco In general, an "equation" is one where we are trying to solve for individual values, but differentiation requires values in an area around the value for $x,$ so in general, if you are trying to solve $f(x)=g(x),$ you cannot differential both sides and get an equation. If, however, for every $x$ in an interval, you have $f(x)=g(x)$, then you can differentiate both sides and still get an equation, potentially more solutions, but containing the solutions in that interval.
$endgroup$
– Thomas Andrews
1 hour ago
$begingroup$
$x=1$ and $x=0$ don't work, either. Differentiating both sides gives you $1=0.$ @Pinteco In general, an "equation" is one where we are trying to solve for individual values, but differentiation requires values in an area around the value for $x,$ so in general, if you are trying to solve $f(x)=g(x),$ you cannot differential both sides and get an equation. If, however, for every $x$ in an interval, you have $f(x)=g(x)$, then you can differentiate both sides and still get an equation, potentially more solutions, but containing the solutions in that interval.
$endgroup$
– Thomas Andrews
1 hour ago
$begingroup$
You can indeed take derivatives, like any other function, as much as you want, on both sides of an equality between objects for which the operation of taking derivatives is defined. The objects being equated in the first equality are not functions, but constant numbers. You could, if you want, tread them as constant functions of some other variable $y$ and then take derivative with respect to $y$. This would give you the true equation $0=0$.
$endgroup$
– user647486
1 hour ago
$begingroup$
You can indeed take derivatives, like any other function, as much as you want, on both sides of an equality between objects for which the operation of taking derivatives is defined. The objects being equated in the first equality are not functions, but constant numbers. You could, if you want, tread them as constant functions of some other variable $y$ and then take derivative with respect to $y$. This would give you the true equation $0=0$.
$endgroup$
– user647486
1 hour ago
$begingroup$
Instead you computed as if taking derivative with respect to $x$. Derivative with respect to $x$ is defined for some functions of $x$, . But that is not an equality between functions of $x$. Equality of functions, by definition, is an equation that is satisfied for all values of $x$.
$endgroup$
– user647486
1 hour ago
$begingroup$
Instead you computed as if taking derivative with respect to $x$. Derivative with respect to $x$ is defined for some functions of $x$, . But that is not an equality between functions of $x$. Equality of functions, by definition, is an equation that is satisfied for all values of $x$.
$endgroup$
– user647486
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
It's important to remember that we can only differentiate functions. When you write the expression
$$
2x^2-x=1
$$
you are no longer dealing with a function. Instead, this expression describes only the solutions $x$ to a given equation. For instance,
$$
f(x) = 2x-x^2
$$
is a function, but $2x-x^2 = 0$ is not.
answered 1 hour ago
rolandcyprolandcyp
1,420313
$endgroup$
add a comment |
$begingroup$
The kicker is that our domain of truth isn't "big enough" to allow it.
From your example, the functions on both sides only agree on $left-frac12,1right$ However, we can't differentiate functions at isolated points of their domains!
On the other hand, consider the equation $$sin x=cos xtan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $fracpi2.$ We can therefore differentiate at all such points, to obtain $$cos x=-sin xtan x+cos xsec^2 x,$$ which one can verify to be true for all such points.
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
$endgroup$
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
add a comment |
$begingroup$
When two functions intersect, they don't have to have the same slopes. For example, $y=x^2, y=x$.![y=x^2,x[1]](https://i.stack.imgur.com/oZkX4.jpg)
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
$endgroup$
add a comment |
$begingroup$
Differentiation isn't an algebraic operation like squaring or addition. Same thing with integration.
You found a counterexample yourself: if you could solve $2x^2-2x=1$ with diff. then you would've gotten $x=-1/2,1$, not (the contradictory statement) $4=0$.
answered 1 hour ago
clathratusclathratus
5,1801438
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's important to remember that we can only differentiate functions. When you write the expression
$$
2x^2-x=1
$$
you are no longer dealing with a function. Instead, this expression describes only the solutions $x$ to a given equation. For instance,
$$
f(x) = 2x-x^2
$$
is a function, but $2x-x^2 = 0$ is not.
answered 1 hour ago
rolandcyprolandcyp
1,420313
$endgroup$
add a comment |
$begingroup$
It's important to remember that we can only differentiate functions. When you write the expression
$$
2x^2-x=1
$$
you are no longer dealing with a function. Instead, this expression describes only the solutions $x$ to a given equation. For instance,
$$
f(x) = 2x-x^2
$$
is a function, but $2x-x^2 = 0$ is not.
answered 1 hour ago
rolandcyprolandcyp
1,420313
$endgroup$
add a comment |
$begingroup$
It's important to remember that we can only differentiate functions. When you write the expression
$$
2x^2-x=1
$$
you are no longer dealing with a function. Instead, this expression describes only the solutions $x$ to a given equation. For instance,
$$
f(x) = 2x-x^2
$$
is a function, but $2x-x^2 = 0$ is not.
answered 1 hour ago
rolandcyprolandcyp
1,420313
$endgroup$
It's important to remember that we can only differentiate functions. When you write the expression
$$
2x^2-x=1
$$
you are no longer dealing with a function. Instead, this expression describes only the solutions $x$ to a given equation. For instance,
$$
f(x) = 2x-x^2
$$
is a function, but $2x-x^2 = 0$ is not.
answered 1 hour ago
rolandcyprolandcyp
1,420313
answered 1 hour ago
rolandcyprolandcyp
1,420313
answered 1 hour ago
rolandcyprolandcyp
1,420313
answered 1 hour ago
rolandcyprolandcyp
1,420313
1,420313
add a comment |
add a comment |
$begingroup$
The kicker is that our domain of truth isn't "big enough" to allow it.
From your example, the functions on both sides only agree on $left-frac12,1right$ However, we can't differentiate functions at isolated points of their domains!
On the other hand, consider the equation $$sin x=cos xtan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $fracpi2.$ We can therefore differentiate at all such points, to obtain $$cos x=-sin xtan x+cos xsec^2 x,$$ which one can verify to be true for all such points.
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
$endgroup$
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
add a comment |
$begingroup$
The kicker is that our domain of truth isn't "big enough" to allow it.
From your example, the functions on both sides only agree on $left-frac12,1right$ However, we can't differentiate functions at isolated points of their domains!
On the other hand, consider the equation $$sin x=cos xtan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $fracpi2.$ We can therefore differentiate at all such points, to obtain $$cos x=-sin xtan x+cos xsec^2 x,$$ which one can verify to be true for all such points.
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
$endgroup$
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
add a comment |
$begingroup$
The kicker is that our domain of truth isn't "big enough" to allow it.
From your example, the functions on both sides only agree on $left-frac12,1right$ However, we can't differentiate functions at isolated points of their domains!
On the other hand, consider the equation $$sin x=cos xtan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $fracpi2.$ We can therefore differentiate at all such points, to obtain $$cos x=-sin xtan x+cos xsec^2 x,$$ which one can verify to be true for all such points.
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
$endgroup$
The kicker is that our domain of truth isn't "big enough" to allow it.
From your example, the functions on both sides only agree on $left-frac12,1right$ However, we can't differentiate functions at isolated points of their domains!
On the other hand, consider the equation $$sin x=cos xtan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $fracpi2.$ We can therefore differentiate at all such points, to obtain $$cos x=-sin xtan x+cos xsec^2 x,$$ which one can verify to be true for all such points.
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
answered 1 hour ago
Cameron BuieCameron Buie
86.1k772161
86.1k772161
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
add a comment |
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
The same thing can be said about indefinite/definite integration?
$endgroup$
– Pinteco
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
$begingroup$
For indefinite integration, how would you define an antiderivative of a function whose domain is a finite set? For definite integration, it turns out to be doable, but we'd end up with $0$ on both sides, rather than what we might expect.
$endgroup$
– Cameron Buie
1 hour ago
add a comment |
$begingroup$
When two functions intersect, they don't have to have the same slopes. For example, $y=x^2, y=x$.
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
$endgroup$
add a comment |
$begingroup$
When two functions intersect, they don't have to have the same slopes. For example, $y=x^2, y=x$.
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
$endgroup$
add a comment |
$begingroup$
When two functions intersect, they don't have to have the same slopes. For example, $y=x^2, y=x$.
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
$endgroup$
When two functions intersect, they don't have to have the same slopes. For example, $y=x^2, y=x$.
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
answered 1 hour ago
Yizhar AmirYizhar Amir
7216
7216
add a comment |
add a comment |
$begingroup$
Differentiation isn't an algebraic operation like squaring or addition. Same thing with integration.
You found a counterexample yourself: if you could solve $2x^2-2x=1$ with diff. then you would've gotten $x=-1/2,1$, not (the contradictory statement) $4=0$.
answered 1 hour ago
clathratusclathratus
5,1801438
$endgroup$
add a comment |
$begingroup$
Differentiation isn't an algebraic operation like squaring or addition. Same thing with integration.
You found a counterexample yourself: if you could solve $2x^2-2x=1$ with diff. then you would've gotten $x=-1/2,1$, not (the contradictory statement) $4=0$.
answered 1 hour ago
clathratusclathratus
5,1801438
$endgroup$
add a comment |
$begingroup$
Differentiation isn't an algebraic operation like squaring or addition. Same thing with integration.
You found a counterexample yourself: if you could solve $2x^2-2x=1$ with diff. then you would've gotten $x=-1/2,1$, not (the contradictory statement) $4=0$.
answered 1 hour ago
clathratusclathratus
5,1801438
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Differentiation isn't an algebraic operation like squaring or addition. Same thing with integration.
You found a counterexample yourself: if you could solve $2x^2-2x=1$ with diff. then you would've gotten $x=-1/2,1$, not (the contradictory statement) $4=0$.
answered 1 hour ago
clathratusclathratus
5,1801438
answered 1 hour ago
clathratusclathratus
5,1801438
answered 1 hour ago
clathratusclathratus
5,1801438
answered 1 hour ago
clathratusclathratus
5,1801438
5,1801438
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add a comment |
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The function $2x^2-2$ is not the same as the function $1$ so it makes no sense to differentiate both sides of that equation the way you have.
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– lulu
1 hour ago
1
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$x=1$ and $x=0$ don't work, either. Differentiating both sides gives you $1=0.$ @Pinteco In general, an "equation" is one where we are trying to solve for individual values, but differentiation requires values in an area around the value for $x,$ so in general, if you are trying to solve $f(x)=g(x),$ you cannot differential both sides and get an equation. If, however, for every $x$ in an interval, you have $f(x)=g(x)$, then you can differentiate both sides and still get an equation, potentially more solutions, but containing the solutions in that interval.
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– Thomas Andrews
1 hour ago
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You can indeed take derivatives, like any other function, as much as you want, on both sides of an equality between objects for which the operation of taking derivatives is defined. The objects being equated in the first equality are not functions, but constant numbers. You could, if you want, tread them as constant functions of some other variable $y$ and then take derivative with respect to $y$. This would give you the true equation $0=0$.
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– user647486
1 hour ago
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Instead you computed as if taking derivative with respect to $x$. Derivative with respect to $x$ is defined for some functions of $x$, . But that is not an equality between functions of $x$. Equality of functions, by definition, is an equation that is satisfied for all values of $x$.
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– user647486
1 hour ago