Find the derivative of the following function.
$$\displaystyle y(x) = (\ln{(2 x)})^{5 x}, \quad x> \frac{1}{2}$$
So the answer given by wolfram seems to be correct:
$$5 \ln^{(5 x-1)}(2 x) (\ln(2 x) \cdot \ln(\ln(2 x))+1)$$
But whence do these $(\ln(2 x) \ln(\ln(2 x))+1)$ come?
Shouldn't there be fractions as derivative of $\ln x = \frac{1}{x}$?
On
Great problem. Okay, it seems you understand that:
$$\frac{d}{dx}a^{x}=a^{x}ln(a)$$
But the Chain Rule gets tricky on this problem because you effectively have a function of two variables:
$$f(x,a)\equiv a^{5x}$$ where $$a(x)\equiv ln(2x)$$ and so the Chain Rule yields: $$\frac{d}{dx}f(x,a(x))=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial a}\frac{da}{dx}$$
The partial derivatives wrt a, x are respectively at constant x, a.
So that applied to your problem as follows:
$$f(x)\equiv(ln(2x))^{5x}$$
$$\frac{df}{dx}=(ln(2x))^{5x}ln(ln(2x))+5x(ln(2x))^{5x-1}\frac{d}{dx}ln(2x)$$
That's how your troubling $\frac{1}{x}$ will get cancelled. I'll let you handle the algebra from there.
Whenever you have a function of the form $y = f(x)^{g(x)}$, it is usually a good idea to use logarithmic differentiation: $$\log y = g(x) \log f(x),$$ hence $$\frac{d}{dx} [ \log y ] = \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[ g(x) \log f(x) \right],$$ and $$\frac{dy}{dx} = f(x)^{g(x)} \left( \frac{g(x)}{f(x)} \frac{df}{dx} + \frac{dg}{dx} \log f(x) \right).$$ So, in your case, $f(x) = \log 2x$ and $g(x) = 5x$, so we have $$\frac{df}{dx} = \frac{1}{2x} \cdot 2 = \frac{1}{x},$$ and $$\frac{dg}{dx} = 5,$$ therefore after minor simplifications, $$\frac{dy}{dx} = (\log 2x)^{5x} \left( \frac{5}{\log 2x} + 5 \log (\log 2x) \right).$$ The only additional step that was done was to put the fractions over a common denominator of $\log 2x$ and factoring: $$\frac{dy}{dx} = 5 (\log 2x)^{5x-1} (1 + (\log 2x) \log (\log 2x)).$$