Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right? 

\begin{array}{l}
f(x) = \sqrt {7x + \sqrt {7x + \sqrt {7x} } } \\
{f^2}(x) = 7x + \sqrt {7x + \sqrt {7x} } \\
{f^2}(x) - 7x = \sqrt {7x + \sqrt {7x} } \\
{\left( {{f^2}(x) - 7x} \right)^2} = 7x + \sqrt {7x} 
\end{array}
Now I'm doing calculus using chain rule:
\begin{array}{l}
2\left( {{f^2}(x) - 7x} \right)\left( {2f(x)f'(x) - 7} \right) = 7\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right)\\
\sqrt {7x + \sqrt {7x} } \left( {2f(x)f'(x) - 7} \right) = \frac{7}{2}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right)\\
2f(x)f'(x) = \frac{7}{2}\frac{1}{{\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 7\\
2f(x)f'(x) = 7\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)\\
f(x)f'(x) = \frac{7}{2}\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)
\end{array}
No needs to factor it out:
\begin{array}{l}
f'(x) = \frac{7}{2}\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)\frac{1}{{f(x)}}\\
f'(x) = \frac{7}{2}\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)\frac{1}{{\sqrt {7x + \sqrt {7x + \sqrt {7x} } } }}
\end{array}
But if somebody wanted to:
$$f'(x) = \left( {\frac{{14\sqrt x  + \sqrt 7 }}{{8\sqrt {7{x^2} + x\sqrt {7x} } }} + \frac{7}{2}} \right)\frac{1}{{\sqrt {7x + \sqrt {7x + \sqrt {7x} } } }}$$
