The derivative represents the instantaneous rate of change, and the rate of change of a product is influenced by both functions' rates and their interaction.

$f(x)g'(x)$ represents the contribution of $f(x)$ to the rate of change, scaled by the rate of change of $g(x)$, and vice versa for $g(x)f'(x)$.

When differentiating a function that is explicitly defined as the product of two other functions.

The derivative of a product involves both the original functions and their derivatives, combined in a specific way.

It's a mnemonic to remember the product rule: (First function) * (Derivative of second) + (Second function) * (Derivative of first).

It is frequently used and tested, often in combination with other differentiation rules.

The instantaneous rate of change of a product of two functions.

It accounts for how the rates of change of two functions combine when they are multiplied together.

It indicates that we are summing the contributions of each function's derivative to the overall derivative of the product.

Because addition is commutative, $f(x)g'(x) + g(x)f'(x)$ is the same as $g(x)f'(x) + f(x)g'(x)$.

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 $g(x)$.

Identify $f(x) = x^2$ and $g(x) = e^x$. Find $f'(x) = 2x$ and $g'(x) = e^x$. Apply the product rule: $y' = x^2e^x + 2xe^x$.

Let $u = x^3 + 1$ and $v = \sin(x)$. Find $u' = 3x^2$ and $v' = \cos(x)$. Use the formula: $f'(x) = (x^3 + 1)\cos(x) + 3x^2\sin(x)$.

Identify $f(x) = 3x^2 - 4x$ and $g(x) = 2x - 1$. Find $f'(x) = 6x - 4$ and $g'(x) = 2$. Apply the Product Rule: $y' = (3x^2-4x)(2) + (2x-1)(6x-4)$.

Identify $u = \sin(x)$ and $v = 3x^2 - 2x + 5$. Find $u' = \cos(x)$ and $v' = 6x - 2$. Use the Product Rule: $f'(x) = \sin(x)(6x - 2) + (3x^2 - 2x + 5)\cos(x)$.

Let $u = e^x$ and $v = \sin(x)$. Find $u' = e^x$ and $v' = \cos(x)$. Apply the Product Rule: $y' = e^x\cos(x) + e^x\sin(x)$.

Identify $u(x) = x$ and $v(x) = \ln(x)$. Find $u'(x) = 1$ and $v'(x) = \frac{1}{x}$. Apply the Product Rule: $f'(x) = x(\frac{1}{x}) + \ln(x)(1) = 1 + \ln(x)$.

Identify $u(x) = x^2$ and $v(x) = \cos(x)$. Find $u'(x) = 2x$ and $v'(x) = -\sin(x)$. Apply the Product Rule: $y' = x^2(-\sin(x)) + \cos(x)(2x) = -x^2\sin(x) + 2x\cos(x)$.

Identify $u(x) = \sqrt{x} = x^{1/2}$ and $v(x) = e^x$. Find $u'(x) = \frac{1}{2}x^{-1/2}$ and $v'(x) = e^x$. Apply the Product Rule: $g'(x) = \sqrt{x}e^x + e^x(\frac{1}{2\sqrt{x}})$.

Identify $u(x) = x^2 + 1$ and $v(x) = x^3 - 1$. Find $u'(x) = 2x$ and $v'(x) = 3x^2$. Apply the Product Rule: $h'(x) = (x^2 + 1)(3x^2) + (x^3 - 1)(2x)$.

Let $u = 2x+3$ and $v = \tan(x)$. Find $u' = 2$ and $v' = \sec^2(x)$. Use the formula: $f'(x) = (2x+3)\sec^2(x) + 2\tan(x)$.