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lgbasallote
 4 years ago
LGBADERIVATIVE:
\[\huge y = e^{k\tan \sqrt x}\]
lgbasallote
 4 years ago
LGBADERIVATIVE: \[\huge y = e^{k\tan \sqrt x}\]

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0im guessing ln again

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

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

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

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

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0this seems complicated :

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

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

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

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

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.4Yes that looks good to me I see no error
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