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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. First, recall the the the product #fg# of the functions #f# and #g# is defined as #(fg)(x)=f(x)g(x)# .
Derivative of the product rule. Related. 2. Derivative using product rule and chain rule. 5. Proof of ...
However, attempting to apply the definition of the derivative directly doesn't seem to work for me. Is there a slick, perhaps intrinsic way, to prove this that doesn't involve working in coordinates? real-analysis
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The product rule for derivatives states that given a function #f(x) = g(x)h(x)#, the derivative of the function is #f'(x) = g'(x)h(x) + g(x)h'(x)# The product rule is used primarily when the function for which one desires the derivative is blatantly the product of two functions, or when the function would be more easily differentiated if looked ...
$\begingroup$ Once you become fluent with the (two-factor) product rule, it would probably help to realize (and justify to yourself) that there's a many-factor product rule: for instance, with four factors, $(fghk)^\prime = f^\prime g h k + f g^\prime h k + f g h^\prime k + f g h k^\prime$.
$\begingroup$ What you are probably missing is the product rule for the derivative of a dot product. $\endgroup$ – Gerry Myerson Commented Jun 17, 2012 at 1:48
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Or you can use the product rule, which works just fine with the cross product: $$ \frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt} $$ Picking a method depends on the problem at hand. For example, the product rule is used to derive Frenet Serret formulas.
$\begingroup$ @BLAZE if you're familiar with first derivatives, simply heed the words in my answer: you need to differentiate once and apply the product rule as needed, and take the resulting function and differentiate again. The product rule is defined for first derivatives, you unfortunately cannot skip steps when differentiating here.