Derivative of the product of two functions

Take a detailed look at the following table and try to deduce the rule of the product:

$f (x)$	$f^{'} (x)$
$x^{2} (3 x + 1)$	$2 x (3 x + 1) + x^{2} (3)$
$4 x (x + 3)$	$4 (x + 3) + 4 x (1)$
$\sin x \cdot c o s x$	$\cos x \cdot \cos x + \sin x \cdot (- \sin x) = \cos^{2} x - \sin^{2} x$
$(A x + B) (A x + B)$	$A (A x + B) + (A x + B) A = 2 A (A x + B)$
$g (x) \cdot h (x)$	?

Have you deduced it? Compare your result with the rule of the product enunciated next.

The derivative of the product of two functions is the derivative of the first one multiplied by the second one plus the first one multiplied by the derivative of the second one.

Mathematically, $f (x) = g (x) h (x) \Rightarrow f^{'} (x) = g^{'} (x) h (x) + g (x) h^{'} (x)$

Some other examples:

Example

$f (x) = 5 x$

We want to derive the previous expression, therefore we need to recognize the functions $g (x)$ and $h (x)$ that should allow us to use the rule of the product. In this case $g (x) = 5$ and $h (x) = x$ . Therefore,

$f^{'} (x) = 0 \cdot x + 5 \cdot 1 = 5$

Example

An example of a function for which we already know the result: $f (x) = x^{2}$

We can take that $g (x) = x$ and $h (x) = x$ and use the rule of the product.

Then, $f^{'} (x) = 1 \cdot x + x \cdot 1 = 2 x$ Obviously the result is the same as the one we already knew.