When two functions are multiplied together — an area whose length and width both change, a revenue that is price times quantity, a polynomial times an exponential — the derivative is not simply the product of the two derivatives. The product rule handles exactly this case, and it is usually the first differentiation rule students meet where intuition alone gives the wrong answer. This calculator applies the rule cleanly and shows the structure of the result, which makes it useful both for checking your own work and for seeing how the two factors share the change between them.
Quick Answer
For h(x) = f(x)·g(x), the derivative is h′(x) = f′(x)·g(x) + f(x)·g′(x). Each factor takes a turn being differentiated while the other is left untouched, and the two results are added.
Why Each Factor Takes a Turn
Picture the product as the area of a rectangle with sides f and g. When x increases a little, both sides grow. The new area picks up a strip along one side (the change in f, times the current g) and a strip along the other side (the change in g, times the current f), plus a tiny corner piece so small it vanishes in the limit. That picture is the whole rule: the derivative counts both strips, which is why there are two terms and why neither term can be dropped.
The formula is symmetric — swapping f and g gives the same answer — but the work is not always symmetric. If one factor is much simpler than the other, differentiate in whichever order keeps the arithmetic light. When the factors are themselves composites or quotients, the product rule still applies at the top level; each f′ or g′ inside it may need its own rule. The rule also extends to three or more factors: differentiate one factor at a time, leave the rest alone, and add all the resulting products together.
How to Use the Product Rule Calculator
- Identify the two factors of your function and decide which you will call f(x) and which g(x).
- Enter the first factor with brackets around it if it contains more than one term, such as (x² − 4).
- Enter the second factor the same way.
- Run the calculation and look at the two-term structure of the answer before any simplification.
- Check each term against your own differentiation of the factors — most errors hide in the factor derivatives, not the rule itself.
- Simplify only at the end, and if the problem asks for a value at a point, substitute into the simplified derivative.
A Worked Example You Can Verify by Expanding
Take h(x) = (x² − 4)(3x + 5). Set f(x) = x² − 4, so f′(x) = 2x, and g(x) = 3x + 5, so g′(x) = 3. The product rule gives h′(x) = 2x·(3x + 5) + (x² − 4)·3 = 6x² + 10x + 3x² − 12 = 9x² + 10x − 12. This example has a satisfying built-in check: expand the original product first and you get 3x³ + 5x² − 12x − 20, and differentiating that term by term also gives 9x² + 10x − 12. When both routes agree, the derivative is certainly right. For products of polynomials, expanding first is always a valid alternative strategy — for products involving eˣ, sin x or ln x, it is not, and the product rule is the only way through.
Slips That Cost Marks with the Product Rule
The overwhelming favourite is writing h′ = f′·g′ — multiplying the two derivatives and dropping the original factors entirely. The second is a bookkeeping error: differentiating both factors in the same term, or differentiating one factor twice. Sign errors creep in when a factor contains a subtraction, because the derivative of (x² − 4) is 2x, but students sometimes carry the −4's ghost into the term. There is also a strategic mistake worth avoiding: applying the product rule to something that is not a product of two functions of x, such as a constant times a function, where simply bringing the constant along is faster and safer. When a product sits inside another function, you will need the Chain Rule Calculator for the outer layer, with the product rule working inside it.
Three Factors and Beyond: Extending the Pattern
The same "one at a time" logic extends to any number of factors, and it is worth seeing once so the pattern is yours. For h(x) = x · eˣ · sin x, differentiate each factor in turn while the other two stand still: the derivative of x contributes eˣ · sin x; the derivative of eˣ contributes x · eˣ · sin x; the derivative of sin x contributes x · eˣ · cos x. Adding the three strips gives h′(x) = eˣ sin x + x·eˣ sin x + x·eˣ cos x, which factors as eˣ · [ sin x · (1 + x) + x · cos x ]. Nothing new was needed — just the discipline of one derivative per term and the patience to take as many terms as there are factors. This generalised form is sometimes called the Leibniz rule, and it explains why the two-factor version has the shape it does. It also gives you a diagnostic: if a product has n factors and your derivative has fewer than n terms (before any lucky simplification), a term has been dropped. Count terms the way you count factors, and the structure of the rule starts checking your work for you.
One more recognition habit is worth building: many functions are products in disguise. The expression x·ln x is a product of x and ln x; so is √x · sin x, and so is any polynomial multiplied by a bracket that cannot be expanded cheaply. Whenever two separate x-dependent factors sit side by side with no operation between them, the product rule is in play — naming the two factors out loud before differentiating is a three-second habit that prevents the commonest wrong answer in this whole topic, the product of the two derivatives.
From Two Factors to the Whole Differentiation Toolkit
Mastering this rule pays off far beyond its own problems, because products appear inside quotients, composites and differential equations. Confirm your answers here, use the Quotient Rule Calculator when the same factors appear as a division instead, and keep the general Derivative Calculator nearby for functions that mix several rules at once.