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If $f$ and $g$ are differentiable, then:
$$\frac{d}{dx}[f(x) \cdot g(x)] = f(x)g'(x) + f'(x)g(x)$$
Or more compactly: $(fg)' = fg' + f'g$
"First times derivative of second, plus second times derivative of first"
$$\frac{d}{dx}[x^2 \sin x] = x^2 \cos x + 2x \sin x$$
$$\frac{d}{dx}[e^x \ln x] = e^x \cdot \frac{1}{x} + e^x \cdot \ln x = e^x\left(\frac{1}{x} + \ln x\right)$$
Consider $f(x) = x$ and $g(x) = x$. Then $fg = x^2$.
If the derivative of a product were the product of derivatives: $(fg)' = f'g' = 1 \cdot 1 = 1$ ❌
But we know $(x^2)' = 2x$ ✓
The product rule gives: $fg' + f'g = x(1) + (1)x = 2x$ ✓
For three functions: $$(fgh)' = f'gh + fg'h + fgh'$$
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