Focuses exclusively on the Product Rule, which is used to find the derivative of a product of two functions, f(x)g(x). This stack includes the formula, conceptual checks, and multiple examples to build proficiency.
It is used to find the derivative of a product of two (or more) functions, where both functions contain the variable of differentiation (e.g., \(x\)).
Is \(\frac{d}{dx}[f(x)g(x)]\) equal to \(f'(x)g'(x)\)?
No. This is a common and critical mistake. The derivative of a product is not the product of the derivatives. You must use the Product Rule.
State the Product Rule for differentiation using Lagrange notation.
If \(h(x) = f(x)g(x)\), then \(h'(x) = f'(x)g(x) + f(x)g'(x)\).
In words: "The derivative of the first function times the second, plus the first function times the derivative of the second."
State the Product Rule for differentiation using Leibniz notation for two functions \(u\) and \(v\).
Find the simplified derivative of \(y = (x^2+1)(x^3-4)\) by expanding the expression for \(y'\).
\(y' = (2x^4 - 8x) + (3x^4 + 3x^2) = 5x^4 + 3x^2 - 8x\).
(Note: While simplifying is good practice, leaving the derivative in its un-simplified product rule form is often acceptable on tests unless specified otherwise).
Find \(\frac{dy}{dx}\) for \(y = (2x+3)\sqrt{x}\) using the Product Rule.
Let \(f(x)=2x+3\) and \(g(x)=x^{1/2}\). Then \(f'(x)=2\) and \(g'(x)=\frac{1}{2}x^{-1/2}\). So:
$$ y' = (2)(x^{1/2}) + (2x+3)(\frac{1}{2}x^{-1/2}) = 2\sqrt{x} + \frac{2x+3}{2\sqrt{x}} $$
How can the Product Rule be extended to find the derivative of three functions, \(h(x) = f(x)g(x)k(x)\)?
By applying the rule iteratively:
$$ h'(x) = f'(x)g(x)k(x) + f(x)g'(x)k(x) + f(x)g(x)k'(x) $$
The derivative is a sum of terms, where in each term, only one of the original functions is differentiated.
Find the equation of the tangent line to the curve \(y = x^2(3x-1)\) at the point where \(x = 2\).
