You've already mastered the power rule, which is great for differentiating simple terms like $x^n$. But what happens when functions get more complex? Specifically, what do we do when we need to find the derivative of two functions multiplied together, or one function divided by another?

It's tempting to think you can just differentiate each part separately. For a product, you might guess the derivative of $f(x)g(x)$ is just $f'(x)g'(x)$. This seems logical, but it's incorrect. Let's see why with a quick example. If $f(x) = x^2$ and $g(x) = x^3$, their product is $x^5$. Using the power rule, the derivative of $x^5$ is $5x^4$. But if we incorrectly differentiate the parts, we get $f'(x)=2x$ and $g'(x)=3x^2$, and their product is $(2x)(3x^2) = 6x^3$. Clearly, $5x^4 \neq 6x^3$. We need a new tool.

When you have two differentiable functions, let's call them $f(x)$ and $g(x)$, multiplied together, you use the Product Rule to find the derivative of their product. The rule looks a bit more involved than the power rule, but it has a consistent structure that's easy to follow with practice.

Let's apply this to a function like $h(x) = (x^2 + 1)(3x - 5)$. Here, we can set $f(x) = x^2 + 1$ and $g(x) = 3x - 5$.

First, we find the derivatives of each part: $f'(x) = 2x$ $g'(x) = 3$

Now, we just plug these pieces into the Product Rule formula: $h'(x) = (2x)(3x - 5) + (x^2 + 1)(3)$

From here, it's just a matter of algebraic simplification. $h'(x) = 6x^2 - 10x + 3x^2 + 3$ $h'(x) = 9x^2 - 10x + 3$

Next, we tackle derivatives of functions that are divided, forming a rational function. Just like with products, we can't simply take the derivative of the numerator and divide it by the derivative of the denominator. We need another specific formula: the Quotient Rule a rule for differentiating fractions of functions.

Notice the similarity to the Product Rule, but with a crucial subtraction and a division by the denominator squared. The order in the numerator matters because of the subtraction. A popular mnemonic for this is "Low D-High minus High D-Low, all over Low-Low."

Let's try it on $h(x) = \frac{2x + 1}{x^2 - 1}$. Here, our "high" function is $f(x) = 2x + 1$ and our "low" function is $g(x) = x^2 - 1$. Their derivatives are: $f'(x) = 2$ $g'(x) = 2x$

Plugging into the Quotient Rule formula: $h'(x) = \frac{(2)(x^2 - 1) - (2x + 1)(2x)}{(x^2 - 1)^2}$

Now we simplify the numerator: $h'(x) = \frac{2x^2 - 2 - (4x^2 + 2x)}{(x^2 - 1)^2}$ $h'(x) = \frac{2x^2 - 2 - 4x^2 - 2x}{(x^2 - 1)^2}$ $h'(x) = \frac{-2x^2 - 2x - 2}{(x^2 - 1)^2}$

It's common to factor out a constant from the numerator if possible, giving a cleaner final answer: $h'(x) = \frac{-2(x^2 + x + 1)}{(x^2 - 1)^2}$

Mastering the Product and Quotient Rules is about being methodical. Write down the separate functions, find their derivatives, and then carefully substitute them into the correct formula. Always be mindful of the signs, especially the subtraction in the Quotient Rule, as that's where most errors happen.