Need help with a second derivative question I am supposed to find the double derivative of
$$ f(x)=5x^2 + 2^x(\ln(3x)) $$
I got that the first derivative is 
$$ 10x+2^x(1/x)+2^x*(\ln(2))(\ln(3x)) $$
I am stuck on the second derivative, I know I am supposed to use the product rule but I am unsure whether I did that rule correctly.
What I ended up with was the following:
$$10+ (2^x \ln(2))(1/x)+(2^x)(-1/x^2)+ (\ln(2))^2) 2^x (1/x) + ((2^x)(\ln(2)^2)) \ln(3x)$$
I know that the notation above looks weird and long, so please tell me if you have issues with the notation
 A: it will make life easier if we call $\ln 3 = b, \ln 2 = a.$ i will find the second derivative of $$y = 2^x \ln(3x) = b 2^x + 2^x \, \ln x \\
y' = ba  2^x + 2^x \frac 1 x + a 2^x \ln x\\
y'' = ba^2 2^x -\frac {2^x}{x^2} + a\frac{2^x}{x} +a^2 2^x \ln x +  a\frac{2^x}{x} =  ba^2 2^x -\frac {2^x}{x^2} + 2a\frac{2^x}{x} +a^2 2^x \ln x$$
A: Introducing$$\begin{gathered}
  u(x) = 5{x^2},u'(x) = 10x,u''(x) = 10 \hfill \\
  p(x) = {2^x},p'(x) = \ln (2){2^x};p''(x) = \ln {(2)^2}{2^x} \hfill \\
  q(x) = \ln (3x),q'(x) = {x^{ - 1}},q''(x) =  - {x^{ - 2}} \hfill \\
  f(x) = u(x) + p(x) \cdot q(x) \hfill \\ 
\end{gathered}$$
and$$f''(x) = u''(x) + p''(x) \cdot q(x) + 2p'(x) \cdot q'(x) + p(x) \cdot q''(x)$$
for second derivative we get:
$$f''(x) = 10 + (\ln {(2)^2} \cdot \ln (3x) \cdot {x^2} + \ln (4) \cdot x - 1) \cdot \frac{{{2^x}}}{{{x^2}}}$$
The second derivative for product rule is a simply application of binomial theorem:
$$\frac{{{d^2}}}{{d{x^2}}}p \cdot q = \sum\limits_{k = 0}^2 {\left( {\begin{array}{*{20}{c}}
  2 \\ 
  k 
\end{array}} \right)}  \cdot \frac{{{d^k}}}{{d{x^k}}}q \cdot \frac{{{d^{2 - k}}}}{{d{x^{2 - k}}}}p$$
