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The derivative f’ ln u defined by f(x)=ln(u(x)) is for all u(x) strictly positive f’(x)=u’(x) / u(x).

Let $u(x)$ be a function of the real variable $x$ such that $u(x) > 0$. The derivative $f’(x)$ of the function $f(x) = \ln(u(x))$ is given by:

$$ \forall x \in \mathbb{R}, \quad f’(x) = \frac{u’(x)}{u(x)} $$

Consider the function $g(x) = \ln(x)$ and the function $h(x) = u(x)$. Then, $f(x) = g(h(x))$. Using the chain rule of differentiation, we have:

$$ f’(x) = g’(h(x)) \cdot h’(x) $$

Using the derivative rule of the natural logarithm function, we have:

$$ g’(x) = \frac{1}{x} $$

Hence:

$$ g’(h(x)) = \frac{1}{u(x)} $$

Using the derivative of the function $u(x)$, we have:

$$ h’(x) = u’(x) $$

Finally, we obtain:

$$ f’(x) = \frac{u’(x)}{u(x)} $$

Therefore, we have:

$$ \forall x \in \mathbb{R}, \quad f’(x) = \frac{u’(x)}{u(x)} $$

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