Proof for Product Rule and Quotient Rule

**stupidguy** · August 23rd 2010, 11:45 AM

My teacher dun give me the Proof for Product Rule and Quotient Rule. She just asked me memorize the technique. Smart bros, pls guide me. I will also ask questions too. thanks

**Chris L T521** · August 23rd 2010, 12:34 PM

Originally Posted by stupidguy

My teacher dun give me the Proof for Product Rule and Quotient Rule. She just asked me memorize the technique. Smart bros, pls guide me. I will also ask questions too. thanks

We resort to the limit definition of the derivative to figure this out.

$\dfrac{d}{dx}[f(x)g(x)]=\lim\limits_{h\to0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}$ .

Now, we add "zero" to the numerator, particularly in the form of $f(x+h)g(x)-f(x+h)g(x)$ .

Thus, we have $\lim\limits_{h\to0}\dfrac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}$ .

Splitting into two different fractions, we get $\lim\limits_{h\to0}\dfrac{f(x+h)g(x+h)-f(x+h)g(x)}{h}+\lim\limits_{h\to0}\dfrac{f(x+h)g(x )-f(x)g(x)}{h}$ .

We now see that we have $\lim\limits_{h\to0}f(x+h)\lim\limits_{h\to0}\dfrac {g(x+h)-g(x)}{h}+g(x)\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=f(x)g^{\prime}(x)+g(x)f^{\prime}(x)$ .

Does this make sense? Do you think you can derive the quotient rule? It can be done in a similar fashion.

**Soroban** · August 23rd 2010, 08:20 PM

Hello, stupidguy!

Here's the Quotient Rule.

We have: . $Q(x) \:=\:\dfrac{f(x)}{g(x)}$
. . and we want: . $\displaystyle Q'(x) \;=\;\lim_{h\to0}\dfrac{Q(x+h) - Q(x)}{h}$

$Q(x+h) - Q(x) \;=\;\dfrac{f(x+h)}{g(x+h)} - \dfrac{f(x)}{g(x)} \;=\;\dfrac{f(x+h)g(x) - f(x)g(x+h)}{g(x)g(x+h)}$

In the numerator, subtract and add $f(x)g(x)$

$Q(x+h)-Q(x) \;=\;\dfrac{f(x+h)(g(x) - f(x)g(x) - f(x)g(x+h) + f(x)g(x)}{g(x)g(x+h)}$

. . . . . . . . . . . . . $=\;\dfrac{g(x)\bigg[f(x+h)-f(x)\bigg] - f(x)\bigg[g(x+h) - g(x)\bigg]}{g(x)g(x+h)}$

Divide by $h\!:$

$\displaystyle \frac{Q(x+h)-Q(x)}{h} \;=\;\frac{g(x)\,\frac{f(x+h)-f(x)}{h} - f(x)\,\frac{g(x+h)-h(x)}{h}}{g(x)g(x+h)}$

Take the limit:

$\displaystyle \lim_{h\to0}\frac{Q(x+h)-Q(x)}{h} \;=\; \lim_{h\to0} \left[\frac{g(x)\,\frac{f(x+h)-f(x)}{h} - f(x)\,\frac{g(x+h)-h(x)}{h}}{g(x)g(x+h)}\right]$

. . $\displaystyle =\; \frac{g(x)\,\overbrace{\lim \frac{f(x+h)-f(x)}{h}}^{\text{This is }f'(x)} \;-\; f(x)\,\overbrace{\lim \frac{g(x+h)-h(x)}{h}}^{\text{This is }g'(x)}}{\underbrace{\lim g(x)g(x+h)}_{\text{This is }[g(x)]^2}}\right]$

Therefore: . $Q'(x) \;=\;\dfrac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$

**stupidguy** · August 24th 2010, 01:30 AM

Originally Posted by Chris L T521

We resort to the limit definition of the derivative to figure this out.

$\dfrac{d}{dx}[f(x)g(x)]=\lim\limits_{h\to0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}$ .

Now, we add "zero" to the numerator, particularly in the form of $f(x+h)g(x)-f(x+h)g(x)$ .

Thus, we have $\lim\limits_{h\to0}\dfrac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}$ .

Splitting into two different fractions, we get $\lim\limits_{h\to0}\dfrac{f(x+h)g(x+h)-f(x+h)g(x)}{h}+\lim\limits_{h\to0}\dfrac{f(x+h)g(x )-f(x)g(x)}{h}$ .

We now see that we have $\lim\limits_{h\to0}f(x+h)\lim\limits_{h\to0}\dfrac {g(x+h)-g(x)}{h}+g(x)\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=f(x)g^{\prime}(x)+g(x)f^{\prime}(x)$ .

Does this make sense? Do you think you can derive the quotient rule? It can be done in a similar fashion.

makes sense. short and sweet. surprisingly my textbook didn't use ur proof to explain.

**stupidguy** · August 24th 2010, 01:35 AM

thanks to both of you. hopefully, I can help others with the knowledge i learnt.

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