Question

LGBADERIVATIVE:

\[\huge y = e^{k\tan \sqrt x}\]

Answer 1

im guessing ln again

Answer 2

Well you can use log diff like I mention before
Or you can use straight chain man

Answer 3

\[\ln y = k\tan \sqrt x\]
now what...

Answer 4

\[y=e^{f(x)} => y'=f'(x)e^{f(x)}\]

Answer 5

i uess i make it
\[\ln y = k\frac{\sin \sqrt x}{\cos \sqrt x}\]

Answer 6

this seems complicated :|

Answer 7

\[y=e^{f(x)}\]
\[\ln(y)=\ln(e^{f(x)})\]
\[\ln(y)=f(x) \cdot \ln(e)\]
\[\ln(y)=f(x) (1)\]
\[\ln(y)=f(x)\]
\[\frac{y'}{y}=f'(x)\]
\[y'=y f'(x)\]
\[y'=e^{f(x)} f'(x)\]

Answer 8

so my problem is that f'(x) now..

Answer 9

\[\frac{d}{dx}\tan(u)=\sec^2(u)\frac{du}{dx}\]

Answer 10

^
what zarkon said :)

Answer 11

uhmm oh yeah...how could i forget *facepalm*

Answer 12

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?

Answer 13

Yes that looks good to me
I see no error

Answer 14

thanks :DDD