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[parent] proof of product rule (Proof)

We begin with two differentiable functions $ f(x)$ and $ g(x)$ and show that their product is differentiable, and that the derivative of the product has the desired form.

By simply calculating, we have for all values of $ x$ in the domain of $ f$ and $ g$ that


$\displaystyle \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}x}}\left[f(x)g(x)\right]$ $\displaystyle =$ $\displaystyle \lim_{h\to0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h}$  
  $\displaystyle =$ $\displaystyle \lim_{h\to0}\frac{f(x+h)g(x+h) + f(x+h)g(x) - f(x+h)g(x) - f(x)g(x)}{h}$  
  $\displaystyle =$ $\displaystyle \lim_{h\to0}\left[f(x+h)\frac{g(x+h)-g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h}\right]$  
  $\displaystyle =$ $\displaystyle \lim_{h\to0}\left[f(x+h)\frac{g(x+h)-g(x)}{h}\right] + \lim_{h\to0}\left[g(x)\frac{f(x+h)-f(x)}{h}\right]$  
  $\displaystyle =$ $\displaystyle f(x)g'(x) + f'(x)g(x).$  

The key argument here is the next to last line, where we have used the fact that both $ f$ and $ g$ are differentiable, hence the limit can be distributed across the sum to give the desired equality.



"proof of product rule" is owned by mathcam. [ full author list (3) | owner history (1) ]
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See Also: derivative, product rule


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Cross-references: equality, sum, limit, line, argument, domain, derivative, differentiable, product, differentiable functions
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This is version 3 of proof of product rule, born on 2002-02-24, modified 2004-10-15.
Object id is 2629, canonical name is ProofOfProductRule.
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Classification:
AMS MSC26A24 (Real functions :: Functions of one variable :: Differentiation : general theory, generalized derivatives, mean-value theorems)

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