The Product Rule The product rule is used when differentiating two functions that are being multiplied together. In some cases it will be possible to simply multiply them out. Example: Differentiate y = x2(x2 + 2x − 3). Here y = x4 + 2x3 − 3x2 and so: However functions like y = 2x(x2 + 1)5 and y = xe3x are either more difficult or impossible to expand and so we need a new technique. The product rule states that for two functions, u and v. If y = uv, then: For our example: y = 2x(x2 − 1)5u = 2xv = (x2 − 1)5Therefore: After factorising: Note: After using the product rule you will normally be able to factorise the derivative and then you can find the stationary points. For our second example: y = xe3x, find the turning point and sketch the graph. u = x v = e3xTherefore: This means there is a stationary point when x = -1/3 (e3x ≠ 0). Also, when x = -1/3, y = -e-1/3 = -0.123 (3sf). By using the second derivative, which we find by using the product rule again, we can determine whether this is a maximum or a minimum. when x = -1/3Therefore there is a minimum at (-1/3, -0.123) To sketch the graph we know that: When x = 0, y = 0 There is a minimum at (-1/3, -0.123) As x → ∞, y → ∞ As x → −∞, y → 0 (and is negative) Therefore the graph looks like this: The Quotient RuleThe quotient rule is actually the product rule in disguise and is used when differentiating a fraction. The quotient rule states that for two functions, u and v, (See if you can use the product rule and the chain rule on y = uv-1 to derive this formula.)Example: DifferentiateSolution: