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? 
 A: Note you can make it slightly less messy by squaring both sides and using implicit differentiation:
$$y^2 = 7x + \sqrt{7x + \sqrt{7x}}$$
$$\implies 2y \dfrac {\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d 7x}{\mathrm dx} + \dfrac {\mathrm d}{\mathrm dx}\left({\sqrt{7x + \sqrt{7x}}}\right) $$
When you're done with the RHS, divide both sides by $2y = 2\sqrt{7x + \sqrt{7x + \sqrt{7x}}}$
A: Build from inside:
$$
\begin{align}
\frac{d}{dx}\sqrt{7x}&=\frac{7}{2\sqrt{7x}}\\
\frac{d}{dx}\sqrt{7x+\sqrt{7x}}&=\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{d}{dx}\sqrt{7x}\right)\\
&=\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{7}{2\sqrt{7x}}\right)\\
\frac{d}{dx}\sqrt{7x+\sqrt{7x+\sqrt{7x}}}&=\frac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\cdot\left(7+\frac{d}{dx}\sqrt{7x+\sqrt{7x}}\right)
\end{align}
$$
where you can plug in the expression for $\frac{d}{dx}\sqrt{7x+\sqrt{7x}}$ found in the second to last line to finish the calculation to end up with
$$
y'=\frac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\cdot\left(7+\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{7}{2\sqrt{7x}}\right)\right)
$$
and now it should be a childs play even finding the derivative of $\sqrt{7x+\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}$ if you feel like it.
A: $$\frac{d}{dx}\sqrt{7x+\sqrt{7x+\sqrt{7x}}}=\frac{7+\frac{7+7/(2\sqrt{7x})}{2\sqrt{7x+\sqrt{7x}}}}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}=\frac{7+14\sqrt{7x}+28\sqrt{7x+\sqrt{7x}}}{8\sqrt{7x}\sqrt{7x+\sqrt{7x}}\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}$$
I hope it is correct.
A: \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:
$$f'(x) = \left( {\frac{{14\sqrt x  + \sqrt 7 }}{{8\sqrt {7{x^2} + x\sqrt {7x} } }} + \frac{7}{2}} \right)\frac{1}{{\sqrt {7x + \sqrt {7x + \sqrt {7x} } } }}$$
A: $$\dfrac{dy}{dx}=\dfrac{d\left( \sqrt{7x+\sqrt{7x+\sqrt{7x}}}\right) }{dx}\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\dfrac{d\left( 7x+\sqrt{7x+\sqrt{7x}}\right) }{dx}\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\dfrac{d\left(\sqrt{7x+\sqrt{7x}}\right) }{dx}\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\dfrac{d\left(7x+\sqrt{7x}\right) }{dx}\right)\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\left(7+\dfrac{d\left(\sqrt{7x}\right) }{dx}\right)\right)\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\left(7+\left(\dfrac{1}{2\sqrt{7x}}\dfrac{d\left(7x\right) }{dx}\right)\right)\right)\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\left(7+\left(\dfrac{7}{2\sqrt{7x}}\right)\right)\right)\right)=\dots$$
