# Thread: Proof for Product Rule and Quotient Rule

1. ## 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

2. 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.

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

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

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

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

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

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

3. Hello, stupidguy!

Here's the Quotient Rule.

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

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

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

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

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

Divide by $\displaystyle h\!:$

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

Take the limit:

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

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

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

4. Originally Posted by Chris L T521
We resort to the limit definition of the derivative to figure this out.

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

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

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

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

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

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.

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