How is the notation $frac{d}{dx} (f^3)(1)$ interpreted?











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How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$




Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?






calculus derivatives







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  • 1




    Where did you get it from? That notation looks very strange.
    – Mike R.
    Nov 12 at 1:27












  • It was in my textbook
    – Is12Prime
    Nov 12 at 1:27










  • Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
    – Rushabh Mehta
    Nov 12 at 1:30






  • 2




    In that case, I'd cautiously vouch for your evaluation.
    – Rushabh Mehta
    Nov 12 at 1:31






  • 1




    I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
    – Sebastian Bechtel
    Nov 12 at 8:19















up vote
4
down vote

favorite
1













How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$




Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?






calculus derivatives







share|cite|improve this question




















  • 1




    Where did you get it from? That notation looks very strange.
    – Mike R.
    Nov 12 at 1:27












  • It was in my textbook
    – Is12Prime
    Nov 12 at 1:27










  • Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
    – Rushabh Mehta
    Nov 12 at 1:30






  • 2




    In that case, I'd cautiously vouch for your evaluation.
    – Rushabh Mehta
    Nov 12 at 1:31






  • 1




    I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
    – Sebastian Bechtel
    Nov 12 at 8:19













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1






How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$




Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?






calculus derivatives







share|cite|improve this question
















How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$




Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?






calculus derivatives




calculus derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 12 at 16:12









user587192

1,410112




1,410112










asked Nov 12 at 1:21









Is12Prime

12119




12119








  • 1




    Where did you get it from? That notation looks very strange.
    – Mike R.
    Nov 12 at 1:27












  • It was in my textbook
    – Is12Prime
    Nov 12 at 1:27










  • Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
    – Rushabh Mehta
    Nov 12 at 1:30






  • 2




    In that case, I'd cautiously vouch for your evaluation.
    – Rushabh Mehta
    Nov 12 at 1:31






  • 1




    I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
    – Sebastian Bechtel
    Nov 12 at 8:19














  • 1




    Where did you get it from? That notation looks very strange.
    – Mike R.
    Nov 12 at 1:27












  • It was in my textbook
    – Is12Prime
    Nov 12 at 1:27










  • Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
    – Rushabh Mehta
    Nov 12 at 1:30






  • 2




    In that case, I'd cautiously vouch for your evaluation.
    – Rushabh Mehta
    Nov 12 at 1:31






  • 1




    I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
    – Sebastian Bechtel
    Nov 12 at 8:19








1




1




Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27






Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27














It was in my textbook
– Is12Prime
Nov 12 at 1:27




It was in my textbook
– Is12Prime
Nov 12 at 1:27












Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30




Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30




2




2




In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31




In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31




1




1




I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19




I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19










1 Answer
1






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up vote
8
down vote



accepted










I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:



$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$






share|cite|improve this answer























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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    8
    down vote



    accepted










    I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:



    $$
    frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
    $$






    share|cite|improve this answer



























      up vote
      8
      down vote



      accepted










      I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:



      $$
      frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
      $$






      share|cite|improve this answer

























        up vote
        8
        down vote



        accepted







        up vote
        8
        down vote



        accepted






        I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:



        $$
        frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
        $$






        share|cite|improve this answer














        I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:



        $$
        frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 12 at 1:39

























        answered Nov 12 at 1:32









        Mike R.

        1,339212




        1,339212






























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