# The Product Rule.

Product Rule is one of the Techniques of Differentiation.

The Product Rule states that:

The Derivative of  product of two Function is the Sum of Derivative of first function multiplied by second function and first function multiplied by derivative of second function.

Mathematically we can write: $\frac{d}{dx}\left(f(x) \times g(x)\right)=g(x) \times \frac{d}{dx}\left(f(x)\right) +f(x) \times \frac{d}{dx}\left(g(x)\right)$

Proof of Product Rule:

If, $h(x)=f(x)\times g(x)$

and ‘a’ is a fixed point Then, $h^|(a)=\displaystyle\lim_{x\to a}\frac{h(x)-h(a)}{x-a}$ $=\displaystyle\lim_{x\to a}\frac{f(x).g(x)-f(a).g(a)}{x-a}$ $=\displaystyle\lim_{x\to a}\frac{f(x).g(x)-f(x).g(a)+f(x).g(a)-f(a).g(a)}{x-a}$ $=\displaystyle\lim_{x\to a}f(x)\frac{g(x)-g(a)}{x-a}+\displaystyle\lim_{x\to a}g(a)\frac{f(x)-f(a)}{x-a}$ $=f(a).g^|(a)+g(a).f^|(a)$

Use of Product Rule:

Find the derivative of: $(3x^2-5x)\times (2x+3)$
Solution: $\frac{d}{dx}\left((3x^2-5x)\times (2x+3)\right) =$ $=(2x+3)\frac{d}{dx}\left(3x^2-5x\right )+(3x^2-5x)\frac{d}{dx}\left(2x+3\right )$ $=(2x+3).(6x-5)+(3x^2-5x).2$ $=18x^2+8x-25$

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