# Algebra Of Derivatives

Rules of Derivatives, Sum Rule, Difference Rule, Product Rule, Quotient Rule, Properties of Derivatives with Solved Examples.

The derivative meaning of function in calculus is the mathematics of continuous change or the rate of change of a quantity concerning another. Consider, for a given function f in x, i.e., f(x), we can find the derivative at every point. It simply means it’s a function f concerning x. The function f changes the input into an output. If we take out the derivative of this function, then the function f will change to f’, df/dx, or f’ (x). As we perform different operations on numbers likewise, we can define algebra as function derivatives, such as sum, difference, product, and quotient.

## Rules of Derivatives

The rules of the derivation include:

• Chain rule formula
• Product rule
• Quotient rule
• Derivatives

Consider, f(x) and g(x) are two functions whose derivatives are defined in a common domain. Now, we can define the following rules of differentiation for the functions f and g.

## Sum rule:

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

The sum rule states that the derivative of a sum is equal to the sum of derivatives.

In mathematics, this means that for f(x)=g(x)+h(x), we can express the derivative of f, f'(x), as f'(x)=g'(x)+h'(x).

For an example, consider a cubic function:

f(x)=Ax3+Bx2+Cx+D.

Where, A=B=C=D= contant.

Now by using three basic properties,

d/dx (c⋅f(x)) =c⋅(df/dx) and d/dx (c)=0, where c represents any constant.

The third Power rule states that, for a quantity xn, d/dx(xn)=nxn-1

If n=1, we would be taking the derivative of x with respect to x, which would inherently be one. Thus dx/dx =1.

Using all four properties, the derivative of a cubic expression is down below:

d/dx f(x)=d/dx [Ax3+Bx2+Cx+D]

=> d/dx (Ax3)+d/dx (Bx2)+d/dx (Cx)+d/dx (D)

=> A (d/dx x3) +B (d/dx x2) +C (d/dx x)+D(d/dx 1)

=> A(3x2) +B(2x) +C (1) +0

=> df/dx=3Ax2+2Bx+C

## Difference rule

The difference of two functions as a derivative is the difference of the derivatives of the functions. Difference rule can be expressed as:

d/dx [f(x)−g(x)]= d/dx( f(x))− d/dx (g(x))

### Product rule:

The Product Rule of Derivative for two functions f(x) and g(x) is given by the formula:

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

## Quotient rule:

A quotient rule example where the denominator is non-zero is as follows:

By considering u = f(x) and v = g(x), the product of these two functions can be written as:

(uv)′ = uv′ + u′v

Also, (u/v)′ = (u′v – uv′)/v²

This is referred to as a Leibnitz rule for differentiating functions’ product and quotient.

## Properties of derivatives

Derivatives can be divided into smaller parts so that the given expressions can be quickly evaluated. In the process of splitting the expressions or functions, the terms are separated based on the operator, such as plus (+), minus (-), or division (/). This can be better understood using the solved examples given below.

## Solved derivative examples

Example 1:

Find the derivative of the function: x3 + (1/x).

Solution: Given: x3 + (1/x)

Let f(x) = x3 and g(x) = 1/x

Using the sum rule of differentiation,

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

d/dx [x3 + (1/x)] = d/dx (x3) + d/dx (1/x)

= 3x2 + (-1/x²)

=  3x2– (1/x²)

Example 2:

Find the derivative of  cos x-sin x.

Solution: Given function is:  cos x-sin x.

Let, f(x) = cos x and g(x) =sin x

Using the difference rule of differentiation,

d/dx [f(x) – g(x)] = d/dx f(x) – d/dx g(x)

d/dx (cos x-sin x) = d/dx (cos x) – d/dx (sin x )

= -sin x-cos x

= -( sin x + cos x )

Example 3:

Solve the derivative for the function: (5x³ – 3x + 1)(x).

Solution: Given: (5x³ – 3x + 1)(x)

Let f(x) = (5x³ – 3x + 1) and g(x) = (x)

Using the product rule of differentiation,

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

= (5x³ – 3x + 1) [d/dx (x)] + (x ) [d/dx (5x³ – 3x + 1)]

= (5x³ – 3x + 1) (1 + 0) + (x )[5(3x2) – 3(1) + 0]

= (5x³ – 3x + 1) + (x)(15×2 – 3)

= 5x³ – 3x + 1 + 15x³ – 3x

= 20x³ – 6x +1

Example 4:

Calculate the derivative of f(x) = 2cot x.

Solution: Given,

f(x) = 2cot x

This can also be written as: f(x) = 2cos x/sin x

Let u(x) = 2cos x and v(x) = sin x

Using quotient rule or Leibnitz rule of quotient,

d/dx f(x) = d/dx [u/v]

= (u/v)′

= (u′v – uv′)/v2

= {[d/dx (2cos x)] sin x –2 cos x[d/dx (sin x)]} / sin2x

= [(-2sin x) sin x – 2cos x (cos x)]/ sin2x

= -[2sin2x + 2cos2x]/ sin2x

= -2/sin2x

= -2cosec2x