What is $frac{d}{df}(frac{df}{dx})$?
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Suppose we have a function $f(x)$. Does $frac{d}{df}big(frac{df}{dx}big)$ exist, and if it does, what is the intuition behind it?
calculus
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Suppose we have a function $f(x)$. Does $frac{d}{df}big(frac{df}{dx}big)$ exist, and if it does, what is the intuition behind it?
calculus
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add a comment |
$begingroup$
Suppose we have a function $f(x)$. Does $frac{d}{df}big(frac{df}{dx}big)$ exist, and if it does, what is the intuition behind it?
calculus
$endgroup$
Suppose we have a function $f(x)$. Does $frac{d}{df}big(frac{df}{dx}big)$ exist, and if it does, what is the intuition behind it?
calculus
calculus
edited Nov 29 '18 at 23:22
amWhy
1
1
asked Nov 29 '18 at 23:19
lithium123lithium123
632315
632315
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2 Answers
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By the chain rule, $frac{mathrm{d}}{mathrm{d}f}left(frac{mathrm{d}f}{mathrm{d}x}right)=frac{mathrm{d}u}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}u}$.
Set $u=x$, for $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Then use $frac{mathrm{d}x}{mathrm{d}f}=frac{1}{frac{mathrm{d}f}{mathrm{d}x}}$ to get $frac{1}{left(frac{mathrm{d}f}{mathrm{d}x}right)}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Or $frac{1}{f'(x)}f''(x)$, if you prefer.
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Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
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– Jam
Nov 29 '18 at 23:46
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By the chain rule it's $f''/f'$ with $g':=dg/dx$. Let's change $x$ to $t$ to discuss an important application in physics. Position $x$ has first two derivatives $v,,a$, so $a=vdv/dx=d(v^2/2)/dx$. This relates Newton's second law to energy conservation.
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2 Answers
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$begingroup$
By the chain rule, $frac{mathrm{d}}{mathrm{d}f}left(frac{mathrm{d}f}{mathrm{d}x}right)=frac{mathrm{d}u}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}u}$.
Set $u=x$, for $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Then use $frac{mathrm{d}x}{mathrm{d}f}=frac{1}{frac{mathrm{d}f}{mathrm{d}x}}$ to get $frac{1}{left(frac{mathrm{d}f}{mathrm{d}x}right)}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Or $frac{1}{f'(x)}f''(x)$, if you prefer.
$endgroup$
$begingroup$
Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
$endgroup$
– Jam
Nov 29 '18 at 23:46
add a comment |
$begingroup$
By the chain rule, $frac{mathrm{d}}{mathrm{d}f}left(frac{mathrm{d}f}{mathrm{d}x}right)=frac{mathrm{d}u}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}u}$.
Set $u=x$, for $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Then use $frac{mathrm{d}x}{mathrm{d}f}=frac{1}{frac{mathrm{d}f}{mathrm{d}x}}$ to get $frac{1}{left(frac{mathrm{d}f}{mathrm{d}x}right)}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Or $frac{1}{f'(x)}f''(x)$, if you prefer.
$endgroup$
$begingroup$
Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
$endgroup$
– Jam
Nov 29 '18 at 23:46
add a comment |
$begingroup$
By the chain rule, $frac{mathrm{d}}{mathrm{d}f}left(frac{mathrm{d}f}{mathrm{d}x}right)=frac{mathrm{d}u}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}u}$.
Set $u=x$, for $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Then use $frac{mathrm{d}x}{mathrm{d}f}=frac{1}{frac{mathrm{d}f}{mathrm{d}x}}$ to get $frac{1}{left(frac{mathrm{d}f}{mathrm{d}x}right)}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Or $frac{1}{f'(x)}f''(x)$, if you prefer.
$endgroup$
By the chain rule, $frac{mathrm{d}}{mathrm{d}f}left(frac{mathrm{d}f}{mathrm{d}x}right)=frac{mathrm{d}u}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}u}$.
Set $u=x$, for $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}left(frac{mathrm{d}f}{mathrm{d}x}right)}{mathrm{d}x}=frac{mathrm{d}x}{mathrm{d}f}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Then use $frac{mathrm{d}x}{mathrm{d}f}=frac{1}{frac{mathrm{d}f}{mathrm{d}x}}$ to get $frac{1}{left(frac{mathrm{d}f}{mathrm{d}x}right)}frac{mathrm{d}^2f}{mathrm{d}x^2}$. Or $frac{1}{f'(x)}f''(x)$, if you prefer.
answered Nov 29 '18 at 23:34
JamJam
4,98521431
4,98521431
$begingroup$
Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
$endgroup$
– Jam
Nov 29 '18 at 23:46
add a comment |
$begingroup$
Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
$endgroup$
– Jam
Nov 29 '18 at 23:46
$begingroup$
Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
$endgroup$
– Jam
Nov 29 '18 at 23:46
$begingroup$
Note that this puts the constraint on whether $frac{mathrm{d}}{mathrm{d}f}frac{mathrm{d}f}{mathrm{d}x}$ can exist since $frac{mathrm{d}f}{mathrm{d}x}$ can't be $0$.
$endgroup$
– Jam
Nov 29 '18 at 23:46
add a comment |
$begingroup$
By the chain rule it's $f''/f'$ with $g':=dg/dx$. Let's change $x$ to $t$ to discuss an important application in physics. Position $x$ has first two derivatives $v,,a$, so $a=vdv/dx=d(v^2/2)/dx$. This relates Newton's second law to energy conservation.
$endgroup$
add a comment |
$begingroup$
By the chain rule it's $f''/f'$ with $g':=dg/dx$. Let's change $x$ to $t$ to discuss an important application in physics. Position $x$ has first two derivatives $v,,a$, so $a=vdv/dx=d(v^2/2)/dx$. This relates Newton's second law to energy conservation.
$endgroup$
add a comment |
$begingroup$
By the chain rule it's $f''/f'$ with $g':=dg/dx$. Let's change $x$ to $t$ to discuss an important application in physics. Position $x$ has first two derivatives $v,,a$, so $a=vdv/dx=d(v^2/2)/dx$. This relates Newton's second law to energy conservation.
$endgroup$
By the chain rule it's $f''/f'$ with $g':=dg/dx$. Let's change $x$ to $t$ to discuss an important application in physics. Position $x$ has first two derivatives $v,,a$, so $a=vdv/dx=d(v^2/2)/dx$. This relates Newton's second law to energy conservation.
answered Nov 29 '18 at 23:24
J.G.J.G.
25.7k22540
25.7k22540
add a comment |
add a comment |
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