Chain rule when applying L'Hopital's rule
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
edited Dec 6 at 11:42
idea
2,00341024
asked Dec 6 at 11:37
Robert
1285
add a comment |
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
edited Dec 6 at 11:42
idea
2,00341024
asked Dec 6 at 11:37
Robert
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
add a comment |
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
edited Dec 6 at 11:42
idea
2,00341024
asked Dec 6 at 11:37
Robert
1285
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
edited Dec 6 at 11:42
idea
2,00341024
asked Dec 6 at 11:37
Robert
1285
edited Dec 6 at 11:42
idea
2,00341024
asked Dec 6 at 11:37
Robert
1285
edited Dec 6 at 11:42
idea
2,00341024
edited Dec 6 at 11:42
idea
2,00341024
edited Dec 6 at 11:42
idea
2,00341024
2,00341024
asked Dec 6 at 11:37
Robert
1285
asked Dec 6 at 11:37
Robert
1285
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 6 at 12:41
gimusi
1
add a comment |
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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.
answered Dec 6 at 12:41
gimusi
1
add a comment |
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.
answered Dec 6 at 12:41
gimusi
1
add a comment |
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.
answered Dec 6 at 12:41
gimusi
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.
answered Dec 6 at 12:41
gimusi
1
edited Dec 6 at 13:46
answered Dec 6 at 12:41
gimusi
1
answered Dec 6 at 12:41
gimusi
1
answered Dec 6 at 12:41
gimusi
1
1
add a comment |
add a comment |
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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