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The derivative of u(x).v(x) is given by : u’(x).v(x) + u(x). v’(x). Let’s prove it using limits.
Let $u(x)$ and $v(x)$ be two functions of the real variable $x$. The derivative $f’(x)$ of the function $f(x) = u(x) \cdot v(x)$ is given by:
$$ \forall x \in \mathbb{R}, \quad f’(x) = u’(x) \cdot v(x) + u(x) \cdot v’(x) $$
Using the definition of the derivative, we have:
$$ \begin{aligned} f’(x) &= \lim_{h\to 0} \frac{u(x+h)v(x+h) - u(x)v(x)}{h} \\ &= \lim_{h\to 0} \frac{u(x+h)v(x+h) - u(x)v(x+h) + u(x)v(x+h) - u(x)v(x)}{h} \\ &= \lim_{h\to 0} \frac{v(x+h)(u(x+h)-u(x))}{h} + \lim_{h\to 0} \frac{u(x)(v(x+h)-v(x))}{h} \\ &= v(x) \cdot \lim_{h\to 0} \frac{u(x+h)-u(x)}{h} + u(x) \cdot \lim_{h\to 0} \frac{v(x+h)-v(x)}{h} \\ &= u’(x) \cdot v(x) + u(x) \cdot v’(x) \end{aligned} $$
Therefore, we have:
$$ \forall x \in \mathbb{R}, \quad f’(x) = u’(x) \cdot v(x) + u(x) \cdot v’(x) $$