# How I do I prove the Product Rule for derivatives?

All we need to do is use the definition of the derivative alongside a simple algebraic trick.

#(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) 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:

Wich give us the product rule

since:

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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