Hey, have I done these correctly? I'm not confidant of the rules at all, especially with b:

a) $\displaystyle y=3sin^2(x)+sec(2x)$
$\displaystyle \frac{dy}{dx}=6cos^2(x)+2sec(2x)tan(2x)$

b)$\displaystyle y=[x+ln(2x)]^3$

With this example, I could solve it easily if not for the $\displaystyle ]^3$ at the end. How would this affect my answer?

2. Originally Posted by Quacky
Hey, have I done these correctly? I'm not confidant of the rules at all, especially with b:

a) $\displaystyle y=3sin^2(x)+sec(2x)$
$\displaystyle \frac{dy}{dx}=6cos^2(x)+2sec(2x)tan(2x)$

you're forgetting to use the chain rule for the first term

b)$\displaystyle y=[x+ln(2x)]^3$

use the chain rule here, also.
...

3. Originally Posted by Quacky
Hey, have I done these correctly? I'm not confidant of the rules at all, especially with b:

a) $\displaystyle y=3sin^2(x)+sec(2x)$
$\displaystyle \frac{dy}{dx}=6cos^2(x)+2sec(2x)tan(2x)$

b)$\displaystyle y=[x+ln(2x)]^3$

With this example, I could solve it easily if not for the $\displaystyle ]^3$ at the end. How would this affect my answer?
$\displaystyle y=3sin^2(x)$
let $\displaystyle u= sin x$
so

$\displaystyle y=3u^2$
then $\displaystyle \frac {dy}{du}=6u \frac {du}{dx}=cosx$

$\displaystyle \frac{dy}{dx}=\frac{dy}{du} \frac {du}{dx}=6ucosx=6sinxcosx$

4. Thanks for the answers, I'll try the second question now.