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.
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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 $$y'=y\times \frac {y'}y$$
Claude Leibovici
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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
$$ f'(x) = \left(\frac{c_{1} x}{c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}}\right)^{c_{6}+c_{7} x+c_{8} \sqrt{c_{9} x}} \left(\left(c_{7}+\frac{c_{8} c_{9}}{2 \sqrt{c_{9} x}}\right) \log \left(\frac{c_{1} x}{c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}}\right)+\frac{c_{5} \left(2 c_{2} \sqrt{c_{5} x}+c_{4} c_{5} x\right) \left(c_{6}+c_{7} x+c_{8} \sqrt{c_{9} x}\right)}{2 (c_{5} x)^{3/2} \left(c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}\right)}\right) $$
dantopa
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