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.

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

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

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

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

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

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

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

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

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

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

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