8
9
3
5
72
27
40
67
8
9
3
5
72
27
40
67
The Product Rule Calculator demonstrates and computes the derivative of a product of two linear functions using the product rule of calculus. Given $$f(x) = ax + b$$ and $$g(x) = cx + d$$, this tool evaluates both functions, their individual derivatives, and applies the product rule formula $$(fg)' = f'g + fg'$$ at any x-value you choose. It is an ideal learning and verification tool for students mastering differentiation rules.
The product rule is one of the fundamental rules of differential calculus. It states that the derivative of a product of two functions is not simply the product of their derivatives — a common misconception among beginning calculus students. Instead, the correct formula is:
$$\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)$$
This rule was first stated by Gottfried Wilhelm Leibniz in his 1684 paper, one of the founding documents of calculus. The product rule can be derived from the limit definition of the derivative by adding and subtracting the term $$f(x+h)g(x)$$ in the difference quotient, then factoring. It generalizes to products of three or more functions: for example, $$(fgh)' = f'gh + fg'h + fgh'$$.
For linear functions $$f(x) = ax + b$$ and $$g(x) = cx + d$$, the product $$f(x) \cdot g(x) = acx^2 + (ad + bc)x + bd$$ is a quadratic. The derivative of this quadratic is $$2acx + (ad + bc)$$. The product rule gives the same result: $$f'g + fg' = a(cx + d) + c(ax + b) = acx + ad + acx + bc = 2acx + ad + bc$$. This agreement confirms the rule and provides a satisfying verification.
While this calculator uses linear functions for clarity, the product rule applies universally to any differentiable functions. It is essential when differentiating expressions like $$x^2 \sin x$$, $$e^x \ln x$$, or $$\sqrt{x} \cos x$$ — any time two non-constant functions are multiplied. In physics, the product rule appears when differentiating momentum $$p = mv$$ when both mass and velocity vary (as in rocket propulsion), or when computing the time derivative of angular momentum.
In engineering, the product rule is used extensively in signal processing when analyzing the product of two time-varying signals, in control theory when linearizing products of state variables, and in economics when differentiating revenue (price times quantity, both functions of another variable). The rule also underlies integration by parts, which is essentially the product rule applied in reverse.
This calculator breaks down each component of the product rule so you can see exactly how $$f'g$$ and $$fg'$$ contribute to the total derivative. Enter the coefficients of two linear functions and an x-value to see the complete computation step by step.
The Product Rule Calculator evaluates the derivative of a product of two linear functions at a specified point.
Step 1: Define the functions.
$$f(x) = ax + b, \quad g(x) = cx + d$$
Step 2: Compute derivatives. Since both functions are linear:
$$f'(x) = a, \quad g'(x) = c$$
Step 3: Evaluate at x₀.
$$f(x_0) = ax_0 + b, \quad g(x_0) = cx_0 + d$$
Step 4: Apply the product rule.
$$(f \cdot g)'(x_0) = f'(x_0) \cdot g(x_0) + f(x_0) \cdot g'(x_0)$$
$$= a(cx_0 + d) + c(ax_0 + b)$$
$$= acx_0 + ad + acx_0 + bc$$
$$= 2acx_0 + ad + bc$$
Verification: The product $$f \cdot g = acx^2 + (ad+bc)x + bd$$ has derivative $$2acx + (ad+bc)$$, which matches the product rule result.
The f(x) and g(x) values show the two functions evaluated at your chosen point. Their product f(x)·g(x) is the value of the combined function.
The f'(x) and g'(x) values are the derivatives of the individual functions. For linear functions, these are simply the coefficients a and c, constant for all x.
The f'(x)·g(x) term represents the contribution to the total derivative from the rate of change of f, weighted by the current value of g. Similarly, f(x)·g'(x) is the contribution from g's rate of change, weighted by f's current value.
The (f·g)' = f'g + fg' is the total derivative of the product, the sum of both contributions. This demonstrates why the product rule requires both terms — neither f'g nor fg' alone gives the correct derivative.
Inputs
Results
f(2) = 6 + 2 = 8. g(2) = 10 − 1 = 9. f'g = 3 × 9 = 27. fg' = 8 × 5 = 40. (fg)' = 27 + 40 = 67. Verification: fg = 15x² + 7x − 2, derivative = 30x + 7, at x=2: 60 + 7 = 67. ✓
Inputs
Results
f(0) = 5. g(0) = 3. f'g = (−2)(3) = −6. fg' = (5)(4) = 20. (fg)' = −6 + 20 = 14. Verification: fg = −8x² + 14x + 15, derivative = −16x + 14, at x=0: 14. ✓
The product rule states that the derivative of a product of two functions equals the first function's derivative times the second, plus the first function times the second's derivative: $$(fg)' = f'g + fg'$$. This rule is necessary because the derivative of a product is generally not equal to the product of the derivatives.
The derivative measures the rate of change. When two quantities both change simultaneously, the total rate of change of their product has two contributions: f changing while g stays momentarily fixed, and g changing while f stays fixed. This is analogous to how the area of a rectangle changes when both length and width vary. Only the product rule captures both contributions correctly.
Yes. For three functions: $$(fgh)' = f'gh + fg'h + fgh'$$. The pattern extends to any number of factors — each function takes a turn being differentiated while the others remain undifferentiated. For $$n$$ functions, the result has $$n$$ terms.
Integration by parts is the product rule in reverse. Starting from $$(fg)' = f'g + fg'$$, integrating both sides gives $$fg = \int f'g\,dx + \int fg'\,dx$$, which rearranges to $$\int fg'\,dx = fg - \int f'g\,dx$$. This is the integration by parts formula, one of the most important integration techniques.
This calculator is designed for linear functions to clearly illustrate the product rule mechanics. The product rule itself applies to any differentiable functions — polynomials, trigonometric, exponential, logarithmic, or any combination. For non-linear functions, the individual derivatives would be more complex, but the rule $$(fg)' = f'g + fg'$$ remains the same.
The product derivative $$(fg)' = f'g + fg' = 0$$ when $$f'g = -fg'$$, i.e., when the two contributions cancel exactly. For linear functions, this happens at $$x = -\frac{ad + bc}{2ac}$$ (provided $$ac \neq 0$$). This is the x-coordinate where the product function $$f \cdot g$$ has a local extremum.
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