Product Rule

In Calculus, the product rule is used to differentiate a function. When a given function is the product of two or more functions, the product rule is used. If the problems are a combination of any two or more functions, then their derivatives can be found using Product Rule. The derivative of a function h(x) will be denoted by D {h(x)} or h'(x).

Product Rule Definition

The product rule is a general rule for the problems which come under the differentiation where one function is multiplied by another function. The derivative of the product of two differentiable functions is equal to the addition of the first function multiplied by the derivative of the second, and the second function multiplied by the derivative of the first function. The function may be exponential, logarithmic function, and so on.

Product Rule Formula

If we have a function y = uv, where u and v are the functions of x. Then, by the use of the product rule, we can easily find out the derivative of y with respect to x, and can be written as:

(dy/dx) = u (dv/dx) + v (du/dx)

The above formula is called the product rule for derivatives or the product rule of differentiation.

In the first term, we have considered u as a constant and for the second term, v as a constant.

Product Rule Proof

Product rule can be proved with the help of limits and by adding, subtracting the one same segment of the function mentioned below:

Let f(x) and g(x) be two functions and h be small increments in the function we get f(x + h) and g(x + h).

Let F(x) = f(x)g(x) and F(x + h) = f(x + h)g(x + h)

Then, the derivative of a function is

By adding and subtracting f(x + h)g(x), we get

By using the definition of a derivative, we get

= f(x + 0) g’ (x) + g(x) f ‘(x)

F'(x) = f(x)g’ (x) + g(x)f ‘(x).

which is the derivative of two functions and is known as the product rule in derivatives.

Product Rule for Different Functions

The product rule for different functions such as derivatives, exponents, logarithmic functions are given below:

Product Rule for Derivatives:

For any two functions, say f(x) and g(x), the product rule is D [f(x) g(x)] = f(x) D[g(x)] + g(x) D[f(x)]

d(uv)/dx = u(dv/dx)+ v(du/dx)

where u and v are two functions

Product Rule for Exponent:

If m and n are the natural numbers, then xⁿ × x^m = x^n+m.

Product rule cannot be used to solve expression of exponent having a different base like 2³* 5⁴ and expressions like (xⁿ)^m. An expression like (xⁿ)^m can be solved only with the help of Power Rule of Exponents where (xⁿ)^m = x^nm.

Product Rule for Logarithm:

For any positive real numbers A and B with the base a

where, a≠ 0, log_aAB = log_aA + log_a B

Product Rule for Partial Derivatives:

If we have a function z = f(x,y) g(x,y) and we want to find out the partial derivative of z, then we use the following formula

Zero Product Rule:

Zero product rule states, the two non zero numbers are only zero if one of them is zero. If a and b are two numbers then ab = 0 only either a = 0 or b = 0.

if (x-1)x = 0, either x – 1 = 0 or x = 0

It means that if x – 1 = 0, then x = 1

Values of x are 0 and 1. They are also called roots of the equation. Mainly used to find the roots of equations, and it works if one side of the equation is zero.

Triple Product Rule:

Triple product rule is a generalization of product rule. If f(x), g(x) and h(x) be three differentiable functions, then the product rule of differentiation can be applied for these three functions as:

D[f(x). g(x). h(x)] = {g(x). h(x)} * D[f(x)] + {f(x). h(x)} * D[g(x)] + {f(x). g(x)} * D[h(x)]

Product Rule Example

Example 1:

Simplify the expression: y= x² × x⁵

Solution:

Given: y= x² × x⁵

We know that the product rule for the exponent is

xⁿ × x^m = x^n+m.

By using the product rule, it can be written as:

y = x² × x⁵ = x²⁺⁵

y = x⁷

Hence, the simplified form of the expression, y= x² × x⁵ is x⁷.

Example 2:

Differentiate y = sin x cos x

Solution:

Given: y = sin x cos x

dy/dx = d(sinx cos x)/dx

While differentiating, it becomes

dy/dx = (sin x) [d(cos x)/dx] + (cos x) [d(sin x)/dx]

Differentiate the terms, dy/dx = sin x (-sin x) + cos x (cos x)

dy/dx = -sin^2.x + cos² x

dy/dx =cos²x – sin²x

By using identity,

dy/dx = cos 2x

Therefore, dy/dx = cos 2x

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