Derivation of the Product Rule for Derivatives
Suppose there are two differentiable functions u(x) and v(x). We need to derive the derivative of their product, i.e., (u(x) \cdot v(x))'.
Step 1: Define increments
First, we consider the increments of the functions u(x) and v(x) at x.
Let \Delta x be a small increment of x, then:
\Delta u = u(x + \Delta x) - u(x)
\Delta v = v(x + \Delta x) - v(x)
Step 2: Increment of the product
Consider the increment of the product u(x) \cdot v(x):
\Delta (u \cdot v) = u(x + \Delta x) \cdot v(x + \Delta x) - u(x) \cdot v(x)
We can transform the expression on the right as follows:
\Delta (u \cdot v) = [u(x + \Delta x) \cdot v(x + \Delta x) - u(x) \cdot v(x + \Delta x)] + [u(x) \cdot v(x + \Delta x) - u(x) \cdot v(x)]
Extracting common factors:
\Delta (u \cdot v) = v(x + \Delta x) \cdot [u(x + \Delta x) - u(x)] + u(x) \cdot [v(x + \Delta x) - v(x)]
That is:
\Delta (u \cdot v) = v(x + \Delta x) \cdot \Delta u + u(x) \cdot \Delta v
Step 3: Derivation
Divide the above expression by \Delta x:
\frac{\Delta (u \cdot v)}{\Delta x} = v(x + \Delta x) \cdot \frac{\Delta u}{\Delta x} + u(x) \cdot \frac{\Delta v}{\Delta x}
As \Delta x approaches 0, v(x + \Delta x) also approaches v(x), thus:
\lim_{\Delta x \to 0} \frac{\Delta (u \cdot v)}{\Delta x} = v(x) \cdot \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} + u(x) \cdot \lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x}
Which means:
(u \cdot v)' = v(x) \cdot u'(x) + u(x) \cdot v'(x)
Conclusion
The product rule for derivatives is:
(u \cdot v)' = u' \cdot v + u \cdot v'