anonymous
  • anonymous
calculate step by step, if existent, via l'Hopital rule: lim-->2 ln(x^2-3)/x^2-4
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

myininaya
  • myininaya
derivative of top/ derivative of bottom
anonymous
  • anonymous
i have difficulty with finding the derivative of top ln(x^2-3)
nubeer
  • nubeer
Hint ln(x+1) [1/(x+1) ] . d/dx (x+1)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
well, what will be the answer if i go step by step?
nubeer
  • nubeer
i dont know.. why dont you solve and put steps here .. if you do something wrong i wil let u know
richyw
  • richyw
\[\frac{d}{dx}\left[\ln\left(x^2-3\right)\right]\]You just need to use the chain rule\[\frac{1}{x^2-3}\cdot\frac{d}{dx}\left(x^2-3\right)=\frac{2x}{x^2-3}\]I already showed you how to check wolfram alpha first for step-by step solutions.
anonymous
  • anonymous
thank you so much... i was confused whether i can use chain or quotient rule in l'hopital or not
richyw
  • richyw
You can use whatever you want to figure out the derivatives. All that matters is the derivative exists (and of course you have an appropriate indeterminate form)

Looking for something else?

Not the answer you are looking for? Search for more explanations.