# proof of product rule

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

 $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)g(x)\right]$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x+h)+f(x+h)g(x)-f(x+h)g(x)-f(x)g(x)}{h}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\left[f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f% (x)}{h}\right]$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\left[f(x+h)\frac{g(x+h)-g(x)}{h}\right]+\lim_{h\to 0% }\left[g(x)\frac{f(x+h)-f(x)}{h}\right]$ $\displaystyle=$ $\displaystyle f(x)g^{\prime}(x)+f^{\prime}(x)g(x).$

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.

Title proof of product rule ProofOfProductRule 2013-03-22 12:28:00 2013-03-22 12:28:00 mathcam (2727) mathcam (2727) 6 mathcam (2727) Proof msc 26A24 Derivative ProductRule