lgbasallote
  • lgbasallote
LGBADERIVATIVE: \[\huge y = e^{k\tan \sqrt x}\]
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lgbasallote
  • lgbasallote
im guessing ln again
freckles
  • freckles
Well you can use log diff like I mention before Or you can use straight chain man
lgbasallote
  • lgbasallote
\[\ln y = k\tan \sqrt x\] now what...

Looking for something else?

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

More answers

freckles
  • freckles
\[y=e^{f(x)} => y'=f'(x)e^{f(x)}\]
lgbasallote
  • lgbasallote
i uess i make it \[\ln y = k\frac{\sin \sqrt x}{\cos \sqrt x}\]
lgbasallote
  • lgbasallote
this seems complicated :|
freckles
  • freckles
\[y=e^{f(x)}\] \[\ln(y)=\ln(e^{f(x)})\] \[\ln(y)=f(x) \cdot \ln(e)\] \[\ln(y)=f(x) (1)\] \[\ln(y)=f(x)\] \[\frac{y'}{y}=f'(x)\] \[y'=y f'(x)\] \[y'=e^{f(x)} f'(x)\]
lgbasallote
  • lgbasallote
so my problem is that f'(x) now..
Zarkon
  • Zarkon
\[\frac{d}{dx}\tan(u)=\sec^2(u)\frac{du}{dx}\]
freckles
  • freckles
^ what zarkon said :)
lgbasallote
  • lgbasallote
uhmm oh yeah...how could i forget *facepalm*
lgbasallote
  • lgbasallote
i need to take a rest hahahaha so it will be \[y' = \frac{e^{k\tan \sqrt x} k\sec^2 \sqrt x}{2\sqrt x}\] right?
freckles
  • freckles
Yes that looks good to me I see no error
lgbasallote
  • lgbasallote
thanks :DDD

Looking for something else?

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