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The derivative of u(x)/v(x) is given by : (u’(x)v(x) - u(x) v’(x))/v^2(x). Let’s prove it using the derivative of an inverse function rule and the product rule for derivatives.

Let $u(x)$ and $v(x)$ be two functions of the real variable $x$ such that $v(x) \neq 0$ and $f(x) = \frac{u(x)}{v(x)}$.

The derivative $f’(x)$ of the function $f(x)$ is:

$$ \forall x \in \mathbb{R}^*, \quad f’(x) = \frac{u’(x) \cdot v(x) - u(x) \cdot v’(x)}{v^2(x)} $$

Expressing $f$ as a product, we have:

$$ f(x)= u(x) \cdot \frac{1}{v(x)} \\ $$

Using the derivative of an inverse function rule, we have:

$$ \left(\frac{1}{v(x)}\right)’ = -\frac{v’(x)}{v^2(x)} $$

Therefore, using the product rule for derivatives, we have:

$$ \begin{aligned} f’(x) &= \left(u(x) \cdot \frac{1}{v(x)}\right)’\\ &= u’(x) \cdot \frac{1}{v(x)} + u(x) \cdot \left(\frac{1}{v(x)}\right)’\\ &= u’(x) \cdot \frac{1}{v(x)} - u(x) \cdot \frac{v’(x)}{v^2(x)}\\ &= \frac{u’(x) \cdot v(x) - u(x) \cdot v’(x)}{v^2(x)} \end{aligned} $$

Therefore:

$$ \forall x \in \mathbb{R}^*, \quad v(x) \neq 0, \quad f’(x) = \frac{u’(x) \cdot v(x) - u(x) \cdot v’(x)}{v^2(x)} $$

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