Calc 1 Section 3-2: Product and Quotient Rules
Welcome back to Professor Baker's Math Class! In this section, we'll be exploring two powerful tools for finding derivatives: the product rule and the quotient rule. These rules allow us to differentiate functions that are expressed as products or quotients of other functions.
The Product Rule
The product rule states that the derivative of a product of two functions is given by:
$$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$$In simpler terms, it's the derivative of the first function times the second function, plus the first function times the derivative of the second function.
Example: Let's find the derivative of $y = (4x^2 + 3)(2x + 5)$.
Here, we can identify $f(x) = 4x^2 + 3$ and $g(x) = 2x + 5$. Then, we find their derivatives:
- $f'(x) = 8x$
- $g'(x) = 2$
Now, applying the product rule:
$$y' = (8x)(2x + 5) + (4x^2 + 3)(2)$$ $$y' = 16x^2 + 40x + 8x^2 + 6$$ $$y' = 24x^2 + 40x + 6$$Another example: Let's differentiate $f(x) = xe^x$. Using the product rule:
- Let $m(x) = x$, so $m'(x) = 1$.
- Let $n(x) = e^x$, so $n'(x) = e^x$.
Then, $f'(x) = (1)(e^x) + (x)(e^x) = e^x + xe^x = e^x(1 + x)$
The Quotient Rule
The quotient rule states that the derivative of a quotient of two functions is given by:
$$\left(\frac{f}{g}\right)' = \frac{gf' - fg'}{g^2}$$Remember this as: (Bottom times derivative of the Top) minus (Top times derivative of the Bottom) all over (Bottom squared)!
Example: Let's find the derivative of $y = \frac{x^2 + x - 2}{x^3 + 6}$.
- $f(x) = x^2 + x - 2$, so $f'(x) = 2x + 1$
- $g(x) = x^3 + 6$, so $g'(x) = 3x^2$
Applying the quotient rule:
$$y' = \frac{(x^3 + 6)(2x + 1) - (x^2 + x - 2)(3x^2)}{(x^3 + 6)^2}$$ $$y' = \frac{(2x^4 + x^3 + 12x + 6) - (3x^4 + 3x^3 - 6x^2)}{(x^3 + 6)^2}$$ $$y' = \frac{-x^4 - 2x^3 + 6x^2 + 12x + 6}{(x^3 + 6)^2}$$Keep practicing, and you'll master these rules in no time! Good luck, and see you in the next section!