Derivative is the rate of change of a function with respect to a variable. The derivatives have so many rules, such as power rule, quotient rule, product rule, and more. They are helpful in solving very complicated problems as well. Derivatives and differentiation do come in higher studies as well with advanced concepts.

Here we will look into what product rule is and how it is used with a formula’s help.

### What is the Product Rule?

When the derivative of two or more functions is to be taken, the product rule is applied. It helps in differentiating between two or more functions in a stated function.

### Leibniz’s Derivative Notation

The derivative f is notated as d/dx*f(x). So when the equation is y = f(x), the derivative is termed as dy/dx. So here, d/dx expresses the differentiation with regard to x. It also indicates the derivative of any given function without using a dependent variable like y² can be denoted as d/dx*y². This is the most used derivative notation as compared to Newton and LaGrange’s notation.

### Derivation of the Formula

Let us take two functions a(x) and b(x). So, the Product rule arrives when you multiply the first function a(x) with the derivative of the second function b(x) plus the derivative of the first function a(x) multiplied by the second function b(x). So,

**(ab)’ = a’b + ab’**

We can prove the derivative product rule by using basic definition of derivative. We can find the increase in the function ab taking that change in argument is Δx :

Δ(ab) = a(x + Δx)b (x + Δx) – a(x)b(x)

Taking into consideration

a(x + Δx) = a(x) + Δa, b(x + Δx) = b(x) + Δb,

Δa and Δb are the increments in the function a and b. Neglecting the brevity of argument of x of function b and a, we can write down the increment Δ(ab) in the form:

Δ(ab) = (a + Δa)( b + Δb) – ab =

~~ab~~+ aΔb + bΔa + ΔaΔb –~~ab~~= aΔb + bΔa + ΔaΔb.

By using properties of limit we can find the derivative of product

(ab)` = limΔx→0 Δ(ab)/Δx

= limΔx→0 (aΔb + bΔa + ΔbΔa)/Δx

= limΔx→0 aΔb /Δx + limΔx→0 bΔa/Δx + limΔx→0 Δa/Δx . limΔx→0 Δb.

The function a does not depend on the increase of Δx. Thus, it is taken outside the limit sign. The same applies to the b. we can calculate the limit limΔx→0 Δb separately

limΔx→0 Δb = limΔx→0 {(Δb/Δx) . Δx }

= limΔx→0 ( Δb )/Δx. limΔx→0 Δx

= b`.0 = 0.

Therefore, the derivative of the product is given by :

(ab)′ = limΔx→0 aΔb /Δx + limΔx→0 ( bΔa )/Δx + limΔx→0 Δa/Δx ⋅ limΔx→0 Δb

= a limΔx→0 Δb /Δx + b limΔx→0 Δa/Δx + limΔx→0 Δa/Δx⋅ limΔx→0 Δb

= ab′ + ba′ + a′⋅0 = a′b + ab′.

From the above formula, it can be easy concluded that derivation of z f(x), where z is a constant :

(z f(x))’= z’ f(x) + z f’(x) = 0.f(x) + z f’(x) = z f’(x)

### Deriving Products of two Functions

Here we will take an example to understand how the product rule is applied to derive the product of two functions.

**Derivative of (x ^{2 }+ x)(3x + 5) = ?**

**Solution:**

So now using the product rule formula, f′(x) = X(x)* Y′(x) + Y(x)* X′(x), we will put the required values.

So here, our first Function X would be (x^{2} + x) while the second function Y would be (3x + 5)

So, Multiply the derivative of the first function to the second derivative and add it to the first section’s derivative multiplied by the second function.

That would look like

(x

^{2}+ x)'(3x + 5) + (x^{2}+ x)(3x + 5)’,= (2x + 1)(3x + 5) + (x

^{2}+ x)(3),Now multiply everything

=6x

^{2}+ 10x + 3x + 5 + 3x^{2}+ 3xSo, now the final outcome is

=9x

^{2}+ 16x + 5

Sometimes students do get confused in calculating product rule. They misunderstand it by calculating the product of the derivatives. But this way the answer would not be correct. Let us make you understand with the same example.

Suppose you calculate the product of the derivates of given function

d/dx(x

^{2}+ x) * (3x + 5),= d/dx (x

^{2 }+ x) d/dx(3x + 5)= (2x + 1) * (3)

= 6x + 3

So here the answer is 6x + 3 which is not the same as 9x

^{2}+ 16x + 5

### Sample Problems on Product Rule

**Problem1: Let y = cos ^{2}x. Differentiate this function by using the**

**product rule.**

**Solution:**

We can represent the function as

y(x) = cosxcosx .

By using product rule,

y′(x)= (cosx cosx)′ = (cosx)′cosx + cosx(cosx)′.

Since (cosx)′ = -sinx, we obtain

y′(x) = -sinxcosx + cosx(-sinx) = -2sinxcosx = -sin2x

**Problem 2: Find the derivative of the function y = e ^{x}sinx**

**Solution:**

By applying product rule

y′(x) = (e

^{x}sinx)′ = (e^{x})′sinx + e^{x}(sinx)′= e

^{x}sinx + e^{x}(cosx)= e

^{x}(sinx + cosx).

**Problem 3: Find the derivative of the function y = xsinx.**

**Solution:**

By the product rule we obtain:

y′(x) = (x sinx)′ = (x)′sinx + x (sinx)′

= sinx + x cosx

**Problem 4: Find the derivative of the function y = x(1 + x).**

**Solution:**

By applying product rule:

y'(x) = {x(1 + x)}’ = x'(1 + x) + x(1 + x)’

= (1 + x) + x(0 + 1)

= 1 + 2x

### Applications in Real World

Derivatives and differentiation do help in solving many real-life problems providing easy solutions. To reach maximum and minimum values of profit, loss, population, cost of material, etc. can be determined with the product rule formula.

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