Question about Product Rule? (basic calculus)

Question

so as you all know the product rule states $\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$

Lets say we have the following: $\frac{d}{dx}[x^2\sin(x)]$

In this case we can use the product rule as such that we take both functions separately:

$f(x)=x^2$ and $g(x)=\sin(x)$

$f'(x)=2x$ and $g'(x)=\cos(x)$

Now this is exactly what I don't get. Like at all. How is it that at the beginning we had this one function "$\frac{d}{dx}[x^2\sin(x)]$" that we had to take the derivative of, and we just split it in half and turned this into two separate functions? It just doesn't make any sense to me unless I'm missing something. Thanks in advance!

Answer 1

What else could one possibly do? I mean, if we consider the function $$f(x) := x^2 \sin(x)$$ and want to find the derivative $f'(x)$, one could apply directly the definition, i.e. calculate the limit $$f'(x) := \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$ However, this is "complicated". So proceeding as usual (i.e. using things we already know), one wants to apply "simpler methods", like for example in this case, we know the derivatives of polynomial functions and power series (the sine function). Thus we want to apply the product rule there.

Answer 2

No you are not missing anything is exactly in this way

$$\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$$

To convince yourself you could take a look to the proof by the definition or also in a not rigorous way observe that

$$\Delta(fg)=(f+\Delta f)(g+ \Delta g)-fg=f\Delta g+g\Delta f+\Delta f\Delta g\approx f\Delta g+g\Delta f$$