How I do I prove the Product Rule for derivatives?

Question

Emily Ammerman · Accepted Answer

All we need to do is use the definition of the derivative alongside a simple algebraic trick. First, recall the the the product #fg# of the functions #f# and #g# is defined as #(fg)(x)=f(x)g(x)#. Therefore, it's derivative is #(fg)^(prime)(x) = lim_(h to 0) ((fg)(x+h)-(fg)(x))/(h) =
lim_(h to 0) (f(x+h)g(x+h)-f(x)g(x))/(h)# Now, note that the expression above is the same as #lim_(h to 0) (f(x+h)g(x+h)+0-f(x)g(x))/(h)# Wich we can rewrite, taking into account that #f(x+h)g(x)-f(x+h)g(x)=0#, as: #lim_(h to 0) 1/h [f(x+h)g(x+h)+(f(x+h)g(x)-f(x+h)g(x))-f(x)g(x)]
= lim_(h to 0) 1/h(f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)])# Using the property that the limit of a sum is the sum of the limits, we get: #lim_(h to 0) f(x+h)(g(x+h)-g(x))/(h) + lim_(h to 0)g(x)(f(x+h)-f(x))/(h)# Wich give us the product rule #(fg)^(prime)(x) = f(x)g^(prime)(x)+g(x)f^(prime)(x),# since: #lim_(h to 0) f(x+h) = f(x),#
#lim_(h to 0)(g(x+h)-g(x))/(h) = g^(prime)(x),#
#lim_(h to 0) g(x)=g(x),#
#lim_(h to 0) (f(x+h)-f(x))/(h) = f^(prime)(x)#

Charles Arocha · Answer

undefined To prove the Product Rule for derivatives, we start with two functions $ f(x) $ and $ g(x) $ that are differentiable at a point $ x $. Let $ h(x) = f(x) \cdot g(x) $. To find $ h'(x) $, we use the definition of the derivative:

$$ h'(x) = \lim_{\Delta x \to 0} \frac{h(x + \Delta x) - h(x)}{\Delta x} $$

Substitute $ h(x) = f(x) \cdot g(x) $:

$$ h'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) \cdot g(x + \Delta x) - f(x) \cdot g(x)}{\Delta x} $$

Now, expand the numerator:

$$ h'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) \cdot g(x + \Delta x) - f(x) \cdot g(x + \Delta x) + f(x) \cdot g(x + \Delta x) - f(x) \cdot g(x)}{\Delta x} $$

Group terms and factor out $ g(x + \Delta x) $ and $ f(x) $:

$$ h'(x) = \lim_{\Delta x \to 0} \left( g(x + \Delta x) \cdot \frac{f(x + \Delta x) - f(x)}{\Delta x} + f(x) \cdot \frac{g(x + \Delta x) - g(x)}{\Delta x} \right) $$

Recognize the definitions of derivatives:

$$ h'(x) = g(x) \cdot f'(x) + f(x) \cdot g'(x) $$

This is the Product Rule for derivatives.