Chain rule when applying L'Hopital's rule












2














I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule







share|cite|improve this question




















  • 2




    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    – Zach
    Dec 6 at 12:14










  • @Zach Oh thanks for the correction!
    – Robert
    Dec 6 at 13:51
















2














I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule







share|cite|improve this question




















  • 2




    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    – Zach
    Dec 6 at 12:14










  • @Zach Oh thanks for the correction!
    – Robert
    Dec 6 at 13:51














2












2








2







I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule







share|cite|improve this question















I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule




derivatives partial-derivative limits-without-lhopital chain-rule






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 at 11:42









idea

2,00341024




2,00341024










asked Dec 6 at 11:37









Robert

1285




1285








  • 2




    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    – Zach
    Dec 6 at 12:14










  • @Zach Oh thanks for the correction!
    – Robert
    Dec 6 at 13:51














  • 2




    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    – Zach
    Dec 6 at 12:14










  • @Zach Oh thanks for the correction!
    – Robert
    Dec 6 at 13:51








2




2




I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
– Zach
Dec 6 at 12:14




I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
– Zach
Dec 6 at 12:14












@Zach Oh thanks for the correction!
– Robert
Dec 6 at 13:51




@Zach Oh thanks for the correction!
– Robert
Dec 6 at 13:51










1 Answer
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oldest

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Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



With reference to your main doubt we have



$$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



and therefore



$$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



Refer also to the related




  • Derivative of a function with respect to another function.






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    Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



    Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



    With reference to your main doubt we have



    $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



    and therefore



    $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



    Refer also to the related




    • Derivative of a function with respect to another function.






    share|cite|improve this answer




























      1














      Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



      Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



      With reference to your main doubt we have



      $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



      and therefore



      $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



      Refer also to the related




      • Derivative of a function with respect to another function.






      share|cite|improve this answer


























        1












        1








        1






        Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



        Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



        With reference to your main doubt we have



        $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



        and therefore



        $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



        Refer also to the related




        • Derivative of a function with respect to another function.






        share|cite|improve this answer














        Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



        Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



        With reference to your main doubt we have



        $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



        and therefore



        $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



        Refer also to the related




        • Derivative of a function with respect to another function.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 at 13:46

























        answered Dec 6 at 12:41









        gimusi

        1




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