lgbasallote
  • lgbasallote
LGBADERIVATIVE: \[\huge y = e^{k\tan \sqrt x}\]
Mathematics
schrodinger
  • schrodinger
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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...

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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

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