In differential calculus, the "Product Rule" says:

• If $f(x)~=~g(x)h(x)$ then
• $f'(x)~=~g'(x)h(x)~+~g(x)h'(x)$
Example:

$x^5$ can be thought of as $x^2x^3,$ so:

• $\frac{d}{dx}~x^5~=~\frac{d}{dx}~(x^2x^3)$
Applying the product rule we get:

• $\frac{d}{dx}~(x^2x^3)~=~(2x^1)(x^3)+(x^2)(3x^2)$
The right hand side then simplifies to $2x^4~+~3x^4$ which is $5x^4,$ as required.
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