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

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

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

With this example, I could solve it easily if not for the $]^3$ 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) $y=3sin^2(x)+sec(2x)$
$\frac{dy}{dx}=6cos^2(x)+2sec(2x)tan(2x)$

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

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

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

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

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

$
y=3u^2
$

then $\frac {dy}{du}=6u
\frac {du}{dx}=cosx$

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

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