Are these derivatives correct (the respective functions involve a square root, fraction and expansion)? Question: Use rules of differentiation to answer the following. There is no need to simplify your answer.


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*If $y = 3x{\sqrt x}$, find $\frac{dy}{dx}$


My working: 
$y = 3x^{1+\frac{1}{2}}$
$y = 3x^{\frac{3}{2}}$
$\frac{dy}{dx} = 3 \left(\frac{3}{2}x^{\frac{3}{2}-1}\right)$
$\frac{dy}{dx} = 3 \left(\frac{3}{2}x^{\frac{1}{2}}\right)$
$\frac{dy}{dx} = \frac{9}{2}x^{\frac{1}{2}}$ (differentiating $x^n$)


*If $f(x) = \frac{2x-1}{5x}$, find $f'(x)$


My working:
$f(x) = \frac{2x-1}{5x}$
$f'(x) = \frac{2(1x^{1-1})-1}{5(1x^{1-1})}$
$f(x) = \frac{1}{5}$ (differentiating $x^n$)


*If $y = (2-3x)^2$, find $\frac{dy}{dx}$


My working:
$y = 9x^2-12x+4$
$f'(x) = 9(2x^{2-1})-12(1x^{1-1})+0$
$f'(x) = 18x-12$ (differentiating $x^n$ and differentiating a constant)

Your work for Question 1 is fine. For Question 2, use the Quotient Rule (you seem to have assumed that the derivative of a fraction is the derivative of the numerator over the derivative of the denominator, which isn't the case):
\begin{align}
f' \left ( x \right ) & = \frac{5x \cdot 2 - \left ( 2x - 1 \right ) \cdot 5}{\left ( 5x \right )^2} = \frac{1}{5x^2}
\end{align}
In the comments you mentioned that your course permits neither the use of the Product Rule nor of the Quotient Rule. In that case, you can split the numerator into two fractions (this is probably a simpler method anyway, but it wasn't the first way that came to mind):
\begin{align}
f' \left ( x \right ) & = \frac{d}{dx} \left ( \frac{2 x}{5 x} - \frac{1}{5x} \right ) = \frac{d}{dx} \left ( \frac{2}{5} - \frac{1}{5x} \right ) = \frac{d}{dx} \left ( - \frac{1}{5x} \right ) = - \frac{1}{5} \left ( - x^{-2} \right ) = \frac{1}{5x^2}
\end{align}
For Question 3, you could have also used the Chain rule, although what you've done is fine. 
