I'm not sure how to derive these:
y = log(2)x^(1/3) (2)=base
and
y = √x(e^3x²)
To your first question:
Rewrite the given equation:
$\displaystyle y = \log_2(x^{\frac13})~\implies~y=\frac13 \cdot \frac1{\ln(2)} \cdot \ln(x)$
You only have to know how $\displaystyle \ln(x)$ is differentiated, all other factors are constant.
Do the second question similarly.
Well, with parentheses to make it clearer: 1/(3(ln 2)x). Yes, that is correct.
In general, the derivative of $\displaystyle e^x$ is $\displaystyle e^x$ while the derivative of $\displaystyle a^x$ with a any positive number, is $\displaystyle a^x ln(a)$. The derivative of $\displaystyle ln(x)$ is $\displaystyle \frac{1}{x}$ while the derivative of $\displaystyle log_a(x)$ with a any positive number is $\displaystyle \frac{1}{x ln(a)}$.
1. I assume that the given function reads:
$\displaystyle y = \sqrt{x} \cdot e^{3x^2}$
2. Re-write so that you have only one power to the base e at the RHS:
$\displaystyle y = e^{3x^2+\frac12 \ln(x)}$
3. Now differentiate using the chain rule:
$\displaystyle y'= e^{3x^2+\frac12 \ln(x)} \cdot \left(6x+\frac1{2x}\right)$
Simplify!
4. If you want to use the product rule and the chain rule you'll get:
$\displaystyle y = \sqrt{x} \cdot e^{3x^2}~\implies~y'=e^{3x^2} \cdot \frac1{2\sqrt{x}} + \sqrt{x} \cdot e^{3x^2} \cdot 6x$
Simplify!