anonymous
  • anonymous
Logarithmic Derivatives?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\(\ \Large f(\theta)=ln(cos\theta) ??\)
nubeer
  • nubeer
ln(x) its derivative is (1/x)d/dx(x)
nubeer
  • nubeer
do the same with cosx

Looking for something else?

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

More answers

anonymous
  • anonymous
So use the Chain Rule?
nubeer
  • nubeer
hmm not so let me draw for u.
nubeer
  • nubeer
|dw:1353048499720:dw| when u have to take derivatives of log functions this is formula u have to remember
anonymous
  • anonymous
?
nubeer
  • nubeer
|dw:1353048585263:dw|
anonymous
  • anonymous
So: |dw:1353048624893:dw| Or do I insert ln(cosx)?
anonymous
  • anonymous
So use the chain rule.
nubeer
  • nubeer
no just cos x
anonymous
  • anonymous
@malical after this step?
anonymous
  • anonymous
Yes. Nubeer said no but his drawing said yes. That d/dx is taking another derivative of just cosx.
anonymous
  • anonymous
Oh okay, that's what I was thinking! Thanks for the help guys!
nubeer
  • nubeer
yeah my bad.. sorry.
anonymous
  • anonymous
That's okay @nubeer. I'm appreciate that you tried to help! ;)

Looking for something else?

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