A method for finding the derivative of a function that is the product of two other functions.

One of the functions being multiplied together.

The other function being multiplied by $f(x)$.

The derivative of the function $f(x)$.

The derivative of the function $g(x)$.

The derivative of a product is NOT the product of the derivatives.

Ensuring accurate calculation of the rate of change of a product of functions.

It can be combined with other rules like the chain rule for more complex functions.

It's a fundamental concept frequently tested in both multiple-choice and free-response questions.

Incorrectly calculating the derivative by only multiplying the derivatives of the individual functions.

The sign of $f(x)g'(x) + g(x)f'(x)$ is positive.

The derivative graph shows the slope of the product function, influenced by the slopes and values of the original functions.

A critical point (local max/min) of the product function $f(x)g(x)$.

By comparing the graph of $f(x)g(x)$ with the combined contributions of $f(x)g'(x)$ and $g(x)f'(x)$.

The derivative, $f'(x)g(x) + f(x)g'(x)$, equals zero at that point.

By observing where $f(x)$ and $g(x)$ are increasing or decreasing, and their respective values.

The net change in the function $f(x)g(x)$ over the given interval.

It is likely to be positive, indicating that $f(x)g(x)$ is also increasing.

It influences the rate at which the slope of the product function changes.

The sign of the derivative will depend on the magnitudes and rates of change of $f(x)$ and $g(x)$.

$\frac{d}{dx}(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$f'(x) = u(x)v'(x) + v(x)u'(x)$

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

$y' = e^x \cos(x) + e^x \sin(x)$

$y' = (3x^2-4x)(2) + (2x-1)(6x-4)$

$f'(x) = \sin(x)(6x-2) + (3x^2-2x+5)\cos(x)$

$\frac{d}{dx}[f(x)g(x)] = f(x) \cdot \frac{d}{dx}[g(x)] + g(x) \cdot \frac{d}{dx}[f(x)]$

$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$