Mixing
Proof of the derivative of $x^n$
$begingroup$
I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative:
begin{align}
(x^n)'&=lim_{h to 0} {(x+h)^n-x^nover h}\
&=lim_{h to 0} {x^n+nx^{n-1}h+{n(n-1)over 2}x^{n-2}h^2+cdots+h^n-x^nover h} \
&=lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+cdots+h^{n-1} right]
end{align}
Because polynomial is continuous for every $x$, we can conclude that $lim_{x_0to 0}(x_0)^n=0$. Therefore
$$lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+dots+h^{n-1} right]= nx^{n-1}$$
Is this proof valid?
calculus proof-verification
edited Jun 22 '15 at 21:45
Cookie
8,786123784
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
$endgroup$
add a comment |
$begingroup$
I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative:
begin{align}
(x^n)'&=lim_{h to 0} {(x+h)^n-x^nover h}\
&=lim_{h to 0} {x^n+nx^{n-1}h+{n(n-1)over 2}x^{n-2}h^2+cdots+h^n-x^nover h} \
&=lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+cdots+h^{n-1} right]
end{align}
Because polynomial is continuous for every $x$, we can conclude that $lim_{x_0to 0}(x_0)^n=0$. Therefore
$$lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+dots+h^{n-1} right]= nx^{n-1}$$
Is this proof valid?
calculus proof-verification
edited Jun 22 '15 at 21:45
Cookie
8,786123784
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
$endgroup$
2
$begingroup$
this looks entirely correct
$endgroup$
– ncmathsadist
Jun 22 '15 at 21:39
4
$begingroup$
Well, unless $n$ is not an integer... but it seems you're set for the simple version.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:40
$begingroup$
The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$lim_{x to a}frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits").
$endgroup$
– Paramanand Singh
Jun 23 '15 at 4:03
add a comment |
$begingroup$
I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative:
begin{align}
(x^n)'&=lim_{h to 0} {(x+h)^n-x^nover h}\
&=lim_{h to 0} {x^n+nx^{n-1}h+{n(n-1)over 2}x^{n-2}h^2+cdots+h^n-x^nover h} \
&=lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+cdots+h^{n-1} right]
end{align}
Because polynomial is continuous for every $x$, we can conclude that $lim_{x_0to 0}(x_0)^n=0$. Therefore
$$lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+dots+h^{n-1} right]= nx^{n-1}$$
Is this proof valid?
calculus proof-verification
edited Jun 22 '15 at 21:45
Cookie
8,786123784
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
$endgroup$
I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative:
begin{align}
(x^n)'&=lim_{h to 0} {(x+h)^n-x^nover h}\
&=lim_{h to 0} {x^n+nx^{n-1}h+{n(n-1)over 2}x^{n-2}h^2+cdots+h^n-x^nover h} \
&=lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+cdots+h^{n-1} right]
end{align}
Because polynomial is continuous for every $x$, we can conclude that $lim_{x_0to 0}(x_0)^n=0$. Therefore
$$lim_{h to 0} left[ nx^{n-1}+{n(n-1)over 2}x^{n-2}h+dots+h^{n-1} right]= nx^{n-1}$$
Is this proof valid?
calculus proof-verification
calculus proof-verification
edited Jun 22 '15 at 21:45
Cookie
8,786123784
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
edited Jun 22 '15 at 21:45
Cookie
8,786123784
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
edited Jun 22 '15 at 21:45
Cookie
8,786123784
edited Jun 22 '15 at 21:45
Cookie
8,786123784
edited Jun 22 '15 at 21:45
Cookie
8,786123784
8,786123784
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
asked Jun 22 '15 at 21:38
gboxgbox
5,48562262
5,48562262
2
$begingroup$
this looks entirely correct
$endgroup$
– ncmathsadist
Jun 22 '15 at 21:39
4
$begingroup$
Well, unless $n$ is not an integer... but it seems you're set for the simple version.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:40
$begingroup$
The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$lim_{x to a}frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits").
$endgroup$
– Paramanand Singh
Jun 23 '15 at 4:03
add a comment |
2
$begingroup$
this looks entirely correct
$endgroup$
– ncmathsadist
Jun 22 '15 at 21:39
4
$begingroup$
Well, unless $n$ is not an integer... but it seems you're set for the simple version.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:40
$begingroup$
The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$lim_{x to a}frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits").
$endgroup$
– Paramanand Singh
Jun 23 '15 at 4:03
2
2
$begingroup$
this looks entirely correct
$endgroup$
– ncmathsadist
Jun 22 '15 at 21:39
$begingroup$
this looks entirely correct
$endgroup$
– ncmathsadist
Jun 22 '15 at 21:39
4
4
$begingroup$
Well, unless $n$ is not an integer... but it seems you're set for the simple version.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:40
$begingroup$
Well, unless $n$ is not an integer... but it seems you're set for the simple version.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:40
$begingroup$
The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$lim_{x to a}frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits").
$endgroup$
– Paramanand Singh
Jun 23 '15 at 4:03
$begingroup$
The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$lim_{x to a}frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits").
$endgroup$
– Paramanand Singh
Jun 23 '15 at 4:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.
I'd rewrite it using this:
For $nge 2$ there exists some polynomial $P(x,h)$ such that
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
$endgroup$
3
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
add a comment |
$begingroup$
The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x rightarrow y'=e^x$.
From the inverse function differentiation rule we find $y=log x rightarrow y'=dfrac{1}{x}$ and (using the chain rule):
$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
1
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.
I'd rewrite it using this:
For $nge 2$ there exists some polynomial $P(x,h)$ such that
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
$endgroup$
3
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
add a comment |
$begingroup$
It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.
I'd rewrite it using this:
For $nge 2$ there exists some polynomial $P(x,h)$ such that
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
$endgroup$
3
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
add a comment |
$begingroup$
It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.
I'd rewrite it using this:
For $nge 2$ there exists some polynomial $P(x,h)$ such that
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
$endgroup$
It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.
I'd rewrite it using this:
For $nge 2$ there exists some polynomial $P(x,h)$ such that
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
edited Jun 22 '15 at 21:48
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
answered Jun 22 '15 at 21:45
ajotatxeajotatxe
53.9k24090
53.9k24090
3
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
add a comment |
3
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
3
3
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
$begingroup$
Is rigor or understanding more important?
$endgroup$
– JacksonFitzsimmons
Jul 3 '15 at 2:15
add a comment |
$begingroup$
The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x rightarrow y'=e^x$.
From the inverse function differentiation rule we find $y=log x rightarrow y'=dfrac{1}{x}$ and (using the chain rule):
$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
1
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
add a comment |
$begingroup$
The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x rightarrow y'=e^x$.
From the inverse function differentiation rule we find $y=log x rightarrow y'=dfrac{1}{x}$ and (using the chain rule):
$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
1
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
add a comment |
$begingroup$
The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x rightarrow y'=e^x$.
From the inverse function differentiation rule we find $y=log x rightarrow y'=dfrac{1}{x}$ and (using the chain rule):
$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x rightarrow y'=e^x$.
From the inverse function differentiation rule we find $y=log x rightarrow y'=dfrac{1}{x}$ and (using the chain rule):
$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
edited Jun 23 '15 at 7:46
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
answered Jun 22 '15 at 22:00
Emilio NovatiEmilio Novati
52.2k43474
52.2k43474
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
1
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
add a comment |
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
1
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
$begingroup$
I think you skipped the part when $n$ is rational. When $n$ is rational the derivative of $x^{n}$ can and should be calculated via algebraical means rather than using the theory of exponential and logarithmic functions.
$endgroup$
– Paramanand Singh
Jun 23 '15 at 3:59
1
1
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
$begingroup$
@Paramanand: This proof is valid for any $a in mathbb{R}$, so also rational. We can give a proof for rational exponents ( positive or negative) without use of the exponential function, but it is more laborious. Note that the derivative of the exponential function can be found very quickly by the definition of derivative using the fact that $e^{x+h}=e^xe^h$.
$endgroup$
– Emilio Novati
Jun 23 '15 at 7:54
add a comment |
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2
$begingroup$
this looks entirely correct
$endgroup$
– ncmathsadist
Jun 22 '15 at 21:39
4
$begingroup$
Well, unless $n$ is not an integer... but it seems you're set for the simple version.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:40
$begingroup$
The proof you give is correct only when $n$ is a positive integer. it is easy to extend the proof for negative integer $n$ as well. There are slight complications when proving the formula for rational $n$. The derivative formula amounts to the following standard limit $$lim_{x to a}frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which I have proved in my post paramanands.blogspot.com/2013/11/… (see "Proof of Standard Limits").
$endgroup$
– Paramanand Singh
Jun 23 '15 at 4:03