Step-by-step derivative of $\left ( \frac{c_1 x}{c_2 x + c_3 + c_4 \sqrt{c_5 x}} \right)^{c_6x + c_7 + c_8 \sqrt{c_9 x}}$ Can someone please walk step by step on how to calculate the derivative
$\left ( \frac{c_1 x}{c_2 x + c_3 + c_4 \sqrt{c_5 x}} \right)^{c_6x + c_7 + c_8 \sqrt{c_9 x}}$
Where the $c_i$ are constants, which are positive.
 A: For this kind of monsters, logarithmic differentiation could be your best friend.
Making the problem more general, consider $$y=\left(\frac{f(x)}{g(x)}\right)^{h(x)}\implies \log(y)=h(x)\left( \log(f(x))-\log(g(x))\right)$$ Differentiate both sides $$\frac {y'}y=h(x) \left(\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}\right)+h'(x)\log\left(\frac{f(x)}{g(x)} \right)$$ Now, use $$y'=y\times \frac {y'}y$$
A: Define 
$$
 f(x) = g(x)^{h(x)} = e^{h(x)\ln g(x)}
$$
Compute
$$
 f'(x) = e^{h(x)\ln g(x)} \left( 
\frac{h(x)g'(x)}{g(x)} + h'(x) \ln g(x)
\right)
$$

Define
$$
\begin{align}
 \alpha(x) &= c_{2} + c_{3} x + c_{4} \sqrt{c_{5}x} \\[2pt]
  g(x) &= \frac{c_{1}x}{\alpha(x)} \\[2pt]
  h(x) &= c_{6} + c_{7} x + c_{8} \sqrt{c_{9} x} 
\end{align}
$$
Derivatives
$$
\begin{align}
  \alpha'(x) &= c_{3} + \frac{c_{4}c_{5}}{2\sqrt{c_{5}x}} \\
  g'(x) &= \frac{c_{1} \left(\alpha( x)- x \alpha'(x)\right)}{\alpha(x)^2} \\[3pt]
  h'(x) &= c_{7}+\frac{c_{8} c_{9}}{2 \sqrt{c_{1} x}}
\end{align}
$$


Final answer
$$
 f'(x) = \left(\frac{c_{1} x}{c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}}\right)^{c_{6}+c_{7} x+c_{8} \sqrt{c_{9} x}} \left(\left(c_{7}+\frac{c_{8} c_{9}}{2 \sqrt{c_{9} x}}\right) \log \left(\frac{c_{1} x}{c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}}\right)+\frac{c_{5} \left(2 c_{2} \sqrt{c_{5} x}+c_{4} c_{5} x\right) \left(c_{6}+c_{7} x+c_{8} \sqrt{c_{9} x}\right)}{2 (c_{5} x)^{3/2} \left(c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}\right)}\right)
$$
