# Proof for Product Rule and Quotient Rule

• August 23rd 2010, 12:45 PM
stupidguy
Proof for Product Rule and Quotient Rule
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
• August 23rd 2010, 01:34 PM
Chris L T521
Quote:

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.
• August 23rd 2010, 09:20 PM
Soroban
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: . $Q'(x) \;=\;\dfrac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$

• August 24th 2010, 02:30 AM
stupidguy
Quote:

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.
• August 24th 2010, 02:35 AM
stupidguy
thanks to both of you. hopefully, I can help others with the knowledge i learnt.