Differentiating $f(x)= sqrt[x]{x}$ from first principles
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An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)
I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?
Note: I am still a beginner, thus it would help greatly if you can include all the steps.
calculus
edited Dec 31 '18 at 15:59
Namaste
1
$endgroup$
add a comment |
$begingroup$
An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)
I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?
Note: I am still a beginner, thus it would help greatly if you can include all the steps.
calculus
edited Dec 31 '18 at 15:59
Namaste
1
$endgroup$
$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34
1
$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12
$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46
add a comment |
$begingroup$
An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)
I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?
Note: I am still a beginner, thus it would help greatly if you can include all the steps.
calculus
edited Dec 31 '18 at 15:59
Namaste
1
$endgroup$
An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)
I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?
Note: I am still a beginner, thus it would help greatly if you can include all the steps.
calculus
calculus
edited Dec 31 '18 at 15:59
Namaste
1
edited Dec 31 '18 at 15:59
Namaste
1
edited Dec 31 '18 at 15:59
Namaste
1
edited Dec 31 '18 at 15:59
Namaste
1
edited Dec 31 '18 at 15:59
Namaste
1
1
asked Dec 31 '18 at 15:26
user630471
$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34
1
$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12
$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46
add a comment |
$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34
1
$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12
$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46
$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34
$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34
1
1
$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12
$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12
$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46
$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$
Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
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I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
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add a comment |
$begingroup$
You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
$$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
$$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$
Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
$endgroup$
add a comment |
$begingroup$
A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$
Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
$endgroup$
add a comment |
$begingroup$
A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$
Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
$endgroup$
A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$
Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
answered Dec 31 '18 at 15:32
cansomeonehelpmeoutcansomeonehelpmeout
7,3193935
7,3193935
add a comment |
add a comment |
$begingroup$
I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
$endgroup$
add a comment |
$begingroup$
I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
$endgroup$
add a comment |
$begingroup$
I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
$endgroup$
I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
answered Dec 31 '18 at 15:31
WuestenfuxWuestenfux
5,3331513
5,3331513
add a comment |
add a comment |
$begingroup$
You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
$$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
$$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
$$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
$$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
$$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
$$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
$endgroup$
You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
$$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
$$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
answered Jan 1 at 10:35
Christian BlatterChristian Blatter
175k8115327
175k8115327
add a comment |
add a comment |
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$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34
1
$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12
$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46