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

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

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

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

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

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

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

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

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

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

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: $f'(x) = (2x+3)\sec^2(x) + 2\tan(x)$.

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