## lgbasallote 3 years ago LGBADERIVATIVE: $\huge y = e^{k\tan \sqrt x}$

1. lgbasallote

im guessing ln again

2. freckles

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

3. lgbasallote

$\ln y = k\tan \sqrt x$ now what...

4. freckles

$y=e^{f(x)} => y'=f'(x)e^{f(x)}$

5. lgbasallote

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

6. lgbasallote

this seems complicated :|

7. freckles

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

8. lgbasallote

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

9. Zarkon

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

10. freckles

^ what zarkon said :)

11. lgbasallote

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

12. 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?

13. freckles

Yes that looks good to me I see no error

14. lgbasallote

thanks :DDD

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