Product Rule in Differentiation

Calculus employs the product rule to differentiate functions. The product rule is employed when a given function is the product of two or more other functions.

This rule, which was discovered by Gottfried Leibniz, allows us to calculate derivatives that we don’t want (or are unable to) multiply rapidly.

With another way of putting it, the product rule allows us to find the derivative of two differentiable functions that are multiplied together by combining our knowledge of both the power rule for derivatives and the sum and difference rule for derivatives.

It can be expressed as follows in simple terms: The second times the derivative of the first multiplied by the derivative of the first multiplied by the second times its own derivative equals the derivative of the second times its own derivative.

Derivation of product rule  

This section introduced us to the product formula, which is used to find the derivatives of two differentiable functions when the product of two differentiable functions is known. In Lagrange’s notation, the product rule can be written as follows for any two functions:

(u v)’ = u’.v+u.v’

Or in Leibniz’s notation as

(d/dx)(u-v) = (du/dx) .v + u dv/dx

Let us now look at the proof of the product rule formula in this section. There are several methods for proving the product rule formula, including the following:

Using the first principle as a guideline

Making use of the chain rule.

Product rule formula 

The product rule formula in Calculus allows us to compute the derivative or evaluate the differentiation of the product of two functions by utilising the product rule formula. The following is the formula for the product rule:

(d/dx)f(x) = (d/dx) {u(x).v'(x)}=[v(x)×u'(x) + u(x)×v'(x)]

Where,

• f(x)= product of differentiable functions u(x) and v(x)

• u(x),v(x)= differentiable functions

• u’(x)=derivative of function u(x)

• v’(x)=derivation of the function v(x)

Zero product rule  

According to the zero product rule, the sum of two non-zero numbers is zero only if one of them is zero. If a and b are two numbers, then ab = 0 only if either a or b is equal to zero. Otherwise, ab = 0.

If (x-1)x = 0, then either x – 1 = 0 or x = 0 is true; otherwise, neither is true.

Essentially, it indicates that if x–1 = 0, then x = 1.

The values of x are between 0 and 1. They are sometimes referred to as the equation’s roots. In most cases, this method is used to discover the roots of equations, and it is effective if one side of the equation is zero.

Derive product rule for differentiation 

All we have to do is use the derivative concept in conjunction with a simple algebraic trick.

To begin, remember that the product fg of the functions f and g is defined as (fg) (x) = f (x) g (x), therefore the derivative is.

(fg)'(x) = lim_h→0 [(fg)(x+h) – (fg)(x)]/h

lim_h→0 [f(x+h) g(x+h) – f(x)g(x)]/h

Now, note that the expression above is the same as,

lim_h→0 [f(x+h) g(x+h) + 0 – f(x)g(x)]/h

Which we can rewrite, taking into account that

f(x+h)g(x) – f(x+h)g(x) = 0 as:

lim_h→0 1/h [f(x+h)g(x+h) + f(x+h)g(x) – f(x+h)g(x)) – f(x)g(x)]

=lim_h→0 1/h (f(x+h)[g(x+h) – g(x)] + g(x)[f(x+h) – f(x)])

Using the property that the limit of a sum is the sum of the limits, we get:

lim_h→0 f(x+h) [g(x+h) – g(x)]/h + lim_h→0 g(x)[f(x+h) – f(x)]/h

Which give us the product rule

(fg)'(x) = f(x)g'(x) + g(x)f'(x),

Since,

lim_h→0 f(x+h) = f(x)

lim_h→0 [g(x+h) – g(x)]/h = g'(x),

lim_h→0 g(x) = g(x),

lim_h→0 [f(x+h) – f(x)]/h = f'(x),

Derivation of product rule formula  

This section introduced us to the product formula, which is used to find the derivatives of two differentiable functions when the product of two differentiable functions is known. In Lagrange’s notation, the product rule can be written as follows for any two functions:

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

Let us see the proof of the product rule formula here. There are different methods to prove the product rule formula, given as,

• Using the first principle

• Using chain rule

Application of product rule in differentiation  

For the derivative of a function of the form h(x) = f(x)g(x), both f(x) and g(x) must be differentiable functions in order for the derivative to be found. With the help of the product rule, we can determine the derivation of a differentiable function h(x) = f(x)g(x) by following the steps outlined below.

• Step 1: note down the values of f(x) and g(x).

• Step 2: find the values of f’(x) and g’(x) and apply the product rule formula, given as:

h'(x) = (d/dx)f(x).g(x) = [g(x)f'(x) + f(x)g'(x)]

Conclusion 

Calculus employs the product rule to differentiate functions. This section introduced us to the product formula, which is used to find the derivatives of two differentiable functions when the product of two differentiable functions is known.The product rule formula in Calculus allows us to compute the derivative or evaluate the differentiation of the product of two functions by utilising the product rule formula.