1. ## 3 derivative questions!

i have no clue how to do this one:
$\displaystyle y = (ln \cdot e^\frac{1}{x} \cdot cos^2 \cdot \sqrt {x})$

this one i tried:
$\displaystyle y = ln (e^3 \cdot \sqrt {x^2 + 2x} \cdot \sqrt {sin2x})$
and the answer i got was:
$\displaystyle y' = \frac {3(x^2+2x) + e(x+1) + e(x^2+2x)(tan2x)}{e(x^2+2x)}$

and this one i have no idea either
$\displaystyle y = log_3 (x^2 + 3x + 5)$

THANKS!!!

2. Originally Posted by qzno
i have no clue how to do this one:
$\displaystyle y = (ln \cdot e^\frac{1}{x} \cdot cos^2 \cdot \sqrt {x})$

this one i tried:
$\displaystyle y = ln (e^3 \cdot \sqrt {x^2 + 2x} \cdot \sqrt {sin2x})$
and the answer i got was:
$\displaystyle y' = \frac {3(x^2+2x) + e(x+1) + e(x^2+2x)(tan2x)}{e(x^2+2x)}$

and this one i have no idea either
$\displaystyle y = log_3 (x^2 + 3x + 5)$

THANKS!!!
$\displaystyle y = \log_3 (x^2 + 3x + 5)$

$\displaystyle 3^y = x^2 + 3x + 5$

$\displaystyle \ln{3^y} = \ln{x^2 + 3x + 5}$

$\displaystyle y \ln{3} = \ln{x^2 + 3x + 5}$

$\displaystyle y = \frac{1}{\ln{3}} \ln{x^2 + 3x + 5}$

Let $\displaystyle u = x^2 + 3x + 5$ so $\displaystyle y = \frac{1}{\ln{3}} \ln{u}$

$\displaystyle \frac{du}{dx} = 2x + 3, \frac{dy}{du} = \frac{1}{\ln{3}} \frac{1}{u} = \frac{1}{(x^2 + 3x + 5)\ln{3}}$

$\displaystyle \frac{dy}{dx} = \frac{2x + 3}{(x^2 + 3x + 5)\ln{3}}$

3. thanks hahah but i figured that one out after

$\displaystyle y = (ln \cdot e^\frac{1}{x} \cdot cos^2 \cdot \sqrt {x})$