What is the first derivative and nth derivative of the following function $ y = \sqrt {2 +\sqrt {3 + \sqrt{x}}}$ What is the first derivative and nth derivative of the following function $ y = \sqrt {2 +\sqrt {3 + \sqrt {x}}}$ 
I think taking the ln for both sides will remove the first square root only?
Could anyone give me a hint ?  
 A: First of all. Use chain rule. I'd recommend learning something new instead of relying on app.(personal opinion)
$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+\sqrt{3+\sqrt{x}}}}.\frac{d\sqrt{3+\sqrt{x}}}{dx}$$
$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+\sqrt{3+\sqrt{x}}}}.\frac{1}{2\sqrt{3+\sqrt{x}}}\frac{d\sqrt{x}}{dx}$$
$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+\sqrt{3+\sqrt{x}}}}.\frac{1}{2\sqrt{3+\sqrt{x}}}\frac{1}{2\sqrt{x}}$$
A: Let $y = \sqrt{2+\sqrt{3+\sqrt x}}$. Step by step we get:
\begin{align}
y^2-2&=\sqrt{3+\sqrt x}\\
y^4-4y^2+1&=\sqrt x\\
(y^4-4y^2+1)^2&=x
\end{align}
Define $f(x) = (x^4-4x^2+1)^2$. By the above, we have $f(y) = x$ and thus $$f'(y)y' = 1 \implies y' = \frac 1{f'(y)}$$
This gives the same formula as in answer by TheDeadLegend.
You can continue to differentiate to get $$f''(y)(y')^2+f'(y)y'' = 0\\
f'''(y)(y')^3+3f''(y)y'y''+f'(y)y'''=0$$
etc. In this way you can find $n$-th derivative recursively. Note that $f$ is a polynomial so its derivatives will vanish eventually (not fast enough to do by hand, in my opinion, but oh well).
Also, you could use Faà di Bruno's formula for higher derivatives of function composition: 
$$\frac{d^n}{dx^n}x=\frac{d^n}{dx^n}f(y)=\sum_{k=1}^nf^{(k)}(y)B_{n,k}(y',y'',\ldots,y^{(n-k+1)})$$ where $B_{n,k}$ are Bell polynomials.
Since, $f$ is (left) inverse of $y$, you might also want to take a look here and here.
A: Alternative approach (same answer) ...
$y^2 = 2+ \sqrt{3+\sqrt{x}}$
$\Rightarrow y^2-2=\sqrt{3 + \sqrt{x}}$
$\Rightarrow y^4-4y^2+4=3+\sqrt{x}$
$\Rightarrow 4y^3\frac{dy}{dx} - 8y\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{8y(y^2-2)\sqrt{x}}$
