Chain rule when applying L'Hopital's rule

Question

I have a very basic question regarding derivation function:

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

$$= \lim_{\omega(t)\to\ 0}(\frac{\frac{d(2)}{d\omega(t)} +[\frac{d(x(t))}{d\omega(t)}\cdot \frac{d\omega(t)}{dt} + \frac{d(\frac{\omega(t)}{dt})}{d\omega(t)}\cdot \frac{dx(t)}{dt}]}{\frac{d\omega(t)}{d\omega(t)}} )$$ So here is my question this $$ \frac{d(x(t))}{d\omega(t)}\cdot \frac{d\omega(t)}{dt}$$ 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.

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

$$\frac{d(x(t))}{d\omega(t)}=\frac{d(x(t))}{dt}\frac{dt}{d\omega(t)}=\frac{\frac{d(x(t))}{dt}}{\frac{d\omega(t)}{dt}}$$

and therefore

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

Refer also to the related

• Derivative of a function with respect to another function.