Product rule is a formula which is used to find the

derivatives of products of two or more

functions. If u(x) and v(x) are functions, continuous in an interval [p, q] with derivatives u

^{1}(a) and v

^{1}(a) respectively at the point x = p, then the derivative of the product y(x) = u(x).v(x) at that point is

y^{1}(p) = u(p) v^{1}(p) + v(p) u^{1}(p)

It is used to find the derivative of two or more functions.

If the equation is y = uvw then

^{dy}⁄_{dx} = uw ^{dv}⁄_{dx} + vw ^{du}⁄_{dx} + uv ^{dw}⁄_{dx}

Example 1: Differentiate y= 3x^{2}.e^{2x}

Answer: Here u = 3x^{2} and v = e^{2x}

^{dy}⁄_{d}x = u ^{dv}⁄_{dx} + v ^{du}⁄_{dx}

^{dy}⁄_{dx} = 3x^{2} . ^{( d )}⁄_{dx} (e^{2x}) + e2x ^{d}⁄_{dx} (3x²)

^{dy}⁄_{dx} = 3x^{2}.(2e^{2x}) + (e^{2x}) (6x)

^{dy}⁄_{dx} = 6x^{2}.e^{2x} + 6x e^{2x}

^{dy}⁄_{dx} = 6xe^{2x} (x+1)

Example 2: Differentiate y = (3x^{3} + 4x^{2} + 2) (5x^{3} +9x)

Answer: If y = uv

^{dy}⁄_{dx} = u ^{dv}⁄_{dx} + v ^{du}⁄_{dx}

Here u = 3x^{3} + 4x^{2}+ 2 and v = 5x^{3} + 9x

^{dy}⁄_{dx} = (3x^{3} + 4x^{2} + 2) d/dx (5x^{3} +9x) + (5x^{3} +9x). d/dx (3x^{3} + 4x^{2} +2)

^{dy}⁄_{dx} = (3x^{3} + 4x^{2} + 2) (15x^{2} + 9) + (5x^{3} +9x). (9x^{2} + 8x)

^{dy}⁄_{dx}=45x^5 + 27x^{3} + 60x^{4} +36x^{2} +30x^{2} +18 +45x^{5} + 40x^{4} +81x^{3} +72x^{2}

^{dy}⁄_{dx}= 90x^{5} +100x^{4} +108x^{3} + 138x^{2} +18