Power Rule, Product Rule, and Quotient Rule

Power Rule Using the Binomial Theorem

Statement:

If $f(x) = x^n$ where $n$ is a constant, then $f'(x) = nx^{n-1}$.

Proof:

To find $f'(x)$, we'll use the limit definition of the derivative: $$ f'(x) = \lim{h \to 0} \frac{f(x+h) - f(x)}{h} Substitute in for $f(x)$: f'(x) = \lim{h \to 0} \frac{(x+h)^n - x^n}{h} $$

Now, let's expand $(x + h)^n$ using the binomial theorem. The binomial theorem states: $$ (x + h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}h^k $$

For our purposes, let's expand the first few terms: $$ (x + h)^n = x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \dots $$

Plugging this into our expression for the derivative, we have: $$ f'(x) = \lim_{h \to 0} \frac{x^n + nx^{n-1}h + \text{terms with } h^2 \text{ and higher} - x^n}{h} $$

Certainly! Let's dive a bit deeper into how we derived the coefficient $n$ for the term $nx^{n-1}$ using the binomial coefficient.

The coefficient of the term $x^{n-1}h$ is given by the binomial coefficient: $$ \binom{n}{1} = \frac{n!}{1!(n-1)!} $$

Breaking down the binomial coefficient:

1. $n!$ is the factorial of $n$, which means $n \times (n-1) \times (n-2) \times \dots \times 1$.
2. $1!$ is simply 1.
3. $(n-1)!$ is the factorial of $(n-1)$, which is $(n-1) \times (n-2) \times \dots \times 1$.

Now, let's compute the value: $$ \binom{n}{1} = \frac{n!}{1!(n-1)!} $$

To simplify, notice that the terms from 1 up to $(n-1)$ in the numerator and denominator will cancel out: $$ \binom{n}{1} = \frac{n \times (n-1) \times (n-2) \times \dots \times 1}{1 \times (n-1) \times (n-2) \times \dots \times 1} = n $$

Thus, we've derived the coefficient $n$ for the term $nx^{n-1}$, which is a crucial part of the power rule.

Combining this with our earlier result, we can see that when differentiating the function $f(x) = x^n$, the term corresponding to $x^{n-1}h$ in the binomial expansion has a coefficient of $n$, resulting in the derivative being $f'(x) = nx^{n-1}$.

The $x^n$ terms cancel out: $$ f'(x) = \lim_{h \to 0} \left( nx^{n-1} + \text{terms with } h \right) $$

Now, since every term in the expansion (after the $nx^{n-1}h$ term) has at least a factor of $h^2$, when we divide by $h$, we're left with terms that still have at least one factor of $h$. As $h \to 0$, all these terms go to 0.

Therefore, the only term that survives is: $$ f'(x) = nx^{n-1} $$

And that completes the proof using the binomial theorem.

Certainly!

This image provides a visual representation of the proof for the product rule of differentiation.

Breaking it down:

1. The top equation defines the derivative of a function $f(x)$:

$$ \frac{df(x)}{dx} = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} $$

This is the fundamental definition of a derivative.

2. The next part of the image sets up the differentiation of the product of two functions, $f(x)g(x)$:

$$ \frac{d}{dx} [f(x)g(x)] = \lim_{{h \to 0}} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} $$

This expression expands the change in the product as $x$ changes by a small amount $h$.

3. The image then breaks this difference quotient into two terms:

$$ = \lim_{{h \to 0}} \left( f(x+h) \frac{g(x+h) - g(x)}{h} + g(x) \frac{f(x+h) - f(x)}{h} \right) $$

The breakdown here is done

cleverly by adding and subtracting the term $f(x+h)g(x)$:

$$ f(x+h)g(x+h) - f(x)g(x) = \left( f(x+h)g(x+h) - f(x+h)g(x) \right) + \left( f(x+h)g(x) - f(x)g(x) \right) $$

This splits the change in the product into the change due to $f$ (holding $g$ constant) and the change due to $g$ (holding $f$ constant).

Finally, when we take the limit as $h \to 0$, each of these terms will lead to the derivatives of $f$ and $g$:

$$ = f'(x)g(x) + f(x)g'(x) $$

This is the product rule: the derivative of the product of two functions is the derivative of the first times the second plus the first times the derivative of the second.

Proof of the Quotient Rule

Given a function of the form $$ u(x) = \frac{f(x)}{g(x)} $$ where both ( f ) and ( g ) are differentiable and ( g(x) \neq 0 ). We want to find the derivative of ( u ).

Step 1: Multiply and divide by the conjugate expression: $$ \frac{u(x+h) - u(x)}{h} = \frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} \times \frac{g(x+h) + g(x)}{g(x+h) + g(x)} $$

Step 2: Simplify: $$ = \frac{f(x+h)g(x) + f(x)g(x+h) - f(x)g(x+h) - f(x+h)g(x)}{h(g(x+h)g(x))} $$

Step 3: Group the terms: $$ = \frac{f(x+h)g(x) - f(x)g(x) + f(x)g(x+h) - f(x+h)g(x)}{h(g(x+h)g(x))} $$

Step 4: Break it up: $$ = \frac{f(x+h) - f(x)}{h} \times \frac{g(x)}{g(x+h)g(x)} + \frac{g(x+h) - g(x)}{h} \times \frac{-f(x)}{g(x+h)g(x)} $$

Step 5: Taking the limit as ( h \to 0 ): $$ \frac{du(x)}{dx} = f'(x)\frac{g(x)}{g^2(x)} - g'(x)\frac{f(x)}{g^2(x)} $$

This simplifies to the quotient rule: $$ \frac{du(x)}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)} $$