Product Rule Calculator
Free Product rule Calculator for calculus. Enter values to get step-by-step solutions with formulas and graphs. Enter your values for instant results.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Product Rule Calculator
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Formula: d/dx[f(x)*g(x)] = f'(x)*g(x) + f(x)*g'(x)
Worked example — h'(2) = 480 | Term 1 (f'g): 192 | Term 2 (fg'): 288 | Second derivative h''(2) = 960
Formula
d/dx[f(x)*g(x)] = f'(x)*g(x) + f(x)*g'(x)
The product rule states that the derivative of a product of two functions equals the derivative of the first times the second, plus the first times the derivative of the second. This extends to n functions, where you differentiate one factor at a time while keeping others unchanged, summing all terms.
Worked Examples
Example 1: Polynomial Product Derivative
Problem:Find the derivative of h(x) = 3x^2 * 2x^3 at x = 2.
Solution:f(x) = 3x^2, f'(x) = 6x g(x) = 2x^3, g'(x) = 6x^2 Product rule: h'(x) = f'(x)*g(x) + f(x)*g'(x) h'(x) = 6x * 2x^3 + 3x^2 * 6x^2 = 12x^4 + 18x^4 = 30x^4 At x = 2: h'(2) = 30 * 16 = 480 Verification: h(x) = 6x^5, h'(x) = 30x^4, h'(2) = 30*16 = 480 (confirmed!) Product value: h(2) = 3(4) * 2(8) = 12 * 16 = 192
Result:h'(2) = 480 | Term 1 (f'g): 192 | Term 2 (fg'): 288 | Second derivative h''(2) = 960
Example 2: Exponential-Polynomial Product
Problem:Differentiate h(x) = 2e^(3x) * 5x^2 at x = 1.
Solution:f(x) = 2e^(3x), f'(x) = 6e^(3x) g(x) = 5x^2, g'(x) = 10x Product rule: h'(x) = 6e^(3x)*5x^2 + 2e^(3x)*10x h'(x) = 30x^2*e^(3x) + 20x*e^(3x) = e^(3x)(30x^2 + 20x) At x = 1: h'(1) = e^3 * (30 + 20) = 20.086 * 50 = 1004.28 Product value: h(1) = 2e^3 * 5 = 10e^3 = 200.855 Logarithmic derivative: h'/h = 3 + 2/1 = 5
Result:h'(1) = 1004.28 | f'g term: 602.57 | fg' term: 401.71 | Log derivative: 5
Frequently Asked Questions
What is the product rule for derivatives and when do you use it?
The product rule states that the derivative of a product of two functions equals the first function times the derivative of the second plus the second function times the derivative of the first: d/dx[f(x)*g(x)] = f'(x)*g(x) + f(x)*g'(x). You use the product rule whenever you need to differentiate the product of two or more functions that both depend on the variable. Common examples include x^2*sin(x), e^x*ln(x), and (x+1)*(x^2-3). The product rule is one of the fundamental differentiation rules alongside the power rule, chain rule, and quotient rule. It extends naturally to products of three or more functions and generalizes to the Leibniz rule for higher-order derivatives.
How do you remember the product rule formula?
Several mnemonics help remember the product rule. The most common is: 'the derivative of the first times the second, plus the first times the derivative of the second.' Some students use 'left d-right plus left right-d' where 'd' means 'take the derivative of.' Another approach: if you think of f*g as two friends, you differentiate one while the other stays the same, then swap who gets differentiated, and add the results. In Leibniz notation, d(fg) = (df)g + f(dg), which looks like distributing the d operator. You can also remember it as analogous to the FOIL method for multiplication, but with derivatives: first-outer plus inner-last becomes first-derivative-second plus first-second-derivative.
How does the product rule extend to three or more functions?
For three functions, the product rule becomes: d/dx[f*g*h] = f'*g*h + f*g'*h + f*g*h'. The pattern is clear: differentiate each function one at a time while keeping the others unchanged, then sum all the terms. For n functions, you get n terms. This can be written compactly as d/dx[f1*f2*...*fn] = sum over i from 1 to n of (product of all fj except fi) times fi'. This generalizes to the Leibniz rule for the n-th derivative of a product, which involves binomial coefficients: the n-th derivative of f*g equals the sum from k=0 to n of C(n,k)*f^(k)*g^(n-k), directly analogous to the binomial theorem for expansion of (a+b)^n.
What is the quotient rule and how does it relate to the product rule?
The quotient rule computes the derivative of f(x)/g(x): d/dx[f/g] = [f'*g - f*g'] / g^2. It can be derived from the product rule by writing f/g as f * g^(-1) and applying the product rule with chain rule: d/dx[f*g^(-1)] = f'*g^(-1) + f*(-1)*g^(-2)*g' = f'/g - f*g'/g^2 = (f'g - fg')/g^2. The quotient rule mnemonic is 'low d-high minus high d-low, square the bottom and away we go.' While mathematically equivalent to the product rule plus chain rule, the quotient rule is a convenient shortcut. Some mathematicians prefer to always use the product rule with negative exponents to avoid memorizing a separate formula.
What is the logarithmic differentiation technique and when is it useful?
Logarithmic differentiation takes the natural log of both sides before differentiating, turning products into sums: if y = f(x)*g(x), then ln(y) = ln(f) + ln(g), so y'/y = f'/f + g'/g, giving y' = y*(f'/f + g'/g). This technique is particularly useful for differentiating functions that are products of many factors (like y = x^2*e^x*sin(x)*ln(x)), functions with variables in both base and exponent (like x^x or x^sin(x)), and expressions where direct application of the product rule would be tedious. For a product of n functions, logarithmic differentiation gives a single formula involving n simple terms rather than n product-rule terms, making it computationally simpler for complex expressions.
How is the product rule used in integration by parts?
Integration by parts is the integral version of the product rule. Starting from d/dx[f*g] = f'g + fg', integrating both sides gives f*g = integral(f'g dx) + integral(fg' dx). Rearranging: integral(fg' dx) = f*g - integral(f'g dx), or equivalently integral(u dv) = uv - integral(v du). This transforms one integral into another that is hopefully simpler. The strategy is to choose u (to differentiate) and dv (to integrate) such that the resulting integral v du is easier than the original. Common applications include integrating x*e^x, x*sin(x), ln(x), and e^x*sin(x). The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) suggests which function to choose as u.
What is elasticity in economics and how does it use the product rule?
Elasticity measures the percentage change in one quantity relative to the percentage change in another, computed as E = (x/y)*(dy/dx). When y is a product of functions (common in economic models like revenue = price * quantity), the product rule is essential for finding dy/dx. For instance, total revenue R = P(Q)*Q where P depends on quantity Q. Then dR/dQ = P'(Q)*Q + P(Q)*1 by the product rule, and marginal revenue analysis requires this derivative. The elasticity of the product equals the sum of the elasticities of the factors, which is the economic interpretation of the logarithmic derivative formula. This connection makes the product rule indispensable in microeconomics, demand analysis, and pricing theory.
What is the Leibniz formula for higher-order derivatives of products?
The Leibniz formula generalizes the product rule to the n-th derivative of a product: d^n/dx^n[f*g] = sum from k=0 to n of C(n,k) * f^(k)(x) * g^(n-k)(x), where C(n,k) is the binomial coefficient n!/(k!(n-k)!) and f^(k) denotes the k-th derivative. For n=1, this gives the standard product rule: f'g + fg'. For n=2: f''g + 2f'g' + fg''. For n=3: f'''g + 3f''g' + 3f'g'' + fg'''. Notice the binomial coefficients 1,2,1 and 1,3,3,1 mirror Pascal triangle, directly paralleling the binomial expansion (a+b)^n. This formula is crucial in differential equations, generating functions, and quantum mechanics where higher-order derivatives of products appear frequently.
How do you handle the product rule with more complex functions like exponentials and trigonometric functions?
When applying the product rule to complex functions, you first identify each factor and compute its individual derivative using appropriate rules (chain rule for compositions, power rule for polynomials, etc.), then combine using the product rule structure. For example, for h(x) = e^(2x) * sin(3x): f = e^(2x) so f' = 2e^(2x) (chain rule), g = sin(3x) so g' = 3cos(3x) (chain rule). Then h' = 2e^(2x)*sin(3x) + e^(2x)*3cos(3x) = e^(2x)[2sin(3x) + 3cos(3x)]. For x^2 * ln(x): f = x^2, f' = 2x, g = ln(x), g' = 1/x, so h' = 2x*ln(x) + x^2*(1/x) = 2x*ln(x) + x. The key is systematic identification of factors and their individual derivatives.
What common mistakes do students make with the product rule?
The most common mistake is assuming the derivative of a product is the product of the derivatives: d/dx[f*g] is NOT equal to f'*g'. This only holds in trivial cases. Another frequent error is forgetting to include both terms of the product rule, especially when one factor seems simple. Students also sometimes confuse the product rule with the chain rule, applying one when they should apply the other. For nested products like f(x)*g(x)*h(x), students may not realize they need to extend the product rule. Sign errors are common when negative coefficients appear. Finally, when combining the product rule with the chain rule (as in differentiating x^2*sin(3x)), students may forget to apply the chain rule to the inner function, missing the factor of 3 in the cos(3x) term.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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