Proof of the derivative of $x^n$












6












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















6












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















6












6








6


3



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 22 '15 at 21:45









Cookie

8,751123784




8,751123784










asked Jun 22 '15 at 21:38









gboxgbox

5,47562262




5,47562262








  • 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




    $begingroup$
    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












2 Answers
2






active

oldest

votes


















9












$begingroup$

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
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$







share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Is rigor or understanding more important?
    $endgroup$
    – JacksonFitzsimmons
    Jul 3 '15 at 2:15



















3












$begingroup$

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):



$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









9












$begingroup$

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
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$







share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Is rigor or understanding more important?
    $endgroup$
    – JacksonFitzsimmons
    Jul 3 '15 at 2:15
















9












$begingroup$

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
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$







share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Is rigor or understanding more important?
    $endgroup$
    – JacksonFitzsimmons
    Jul 3 '15 at 2:15














9












9








9





$begingroup$

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
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$







share|cite|improve this answer











$endgroup$



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
$$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 22 '15 at 21:48

























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














  • 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











3












$begingroup$

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):



$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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):



$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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):



$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$






share|cite|improve this answer











$endgroup$



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):



$$
y=x^a =e^{a log x} rightarrow y'=e^{a log x} (alog x)'=e^{a log x} dfrac{a}{x}=ax^{a-1}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 23 '15 at 7:46

























answered Jun 22 '15 at 22:00









Emilio NovatiEmilio Novati

52.1k43474




52.1k43474












  • $begingroup$
    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






  • 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




















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