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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    3












    $begingroup$


    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







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    $endgroup$















      3












      3








      3





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      edited Nov 29 '18 at 23:22









      amWhy

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      asked Nov 29 '18 at 23:19









      lithium123lithium123

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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.






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          • $begingroup$
            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



















          2












          $begingroup$

          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.






          share|cite|improve this answer









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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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
















            3












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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














            3












            3








            3





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • $begingroup$
              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











            2












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 23:24









                J.G.J.G.

                25.7k22540




                25.7k22540






























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