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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
im guessing ln again
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.4
Well you can use log diff like I mention before Or you can use straight chain man
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
\[\ln y = k\tan \sqrt x\] now what...
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.4
\[y=e^{f(x)} => y'=f'(x)e^{f(x)}\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i uess i make it \[\ln y = k\frac{\sin \sqrt x}{\cos \sqrt x}\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
this seems complicated :
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.4
\[y=e^{f(x)}\] \[\ln(y)=\ln(e^{f(x)})\] \[\ln(y)=f(x) \cdot \ln(e)\] \[\ln(y)=f(x) (1)\] \[\ln(y)=f(x)\] \[\frac{y'}{y}=f'(x)\] \[y'=y f'(x)\] \[y'=e^{f(x)} f'(x)\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so my problem is that f'(x) now..
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{d}{dx}\tan(u)=\sec^2(u)\frac{du}{dx}\]
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.4
^ what zarkon said :)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
uhmm oh yeah...how could i forget *facepalm*
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.4
Yes that looks good to me I see no error
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
thanks :DDD
 2 years ago
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