# Algebra of Derivative of Functions

The derivative of a function in calculus is the rate of change of a quantity with respect to another. Also, evaluating the derivative of a given function at a certain point involves effective use of various rules, limits are subjected to. For a given function f in x, i.e. f(x) we can find the derivative at every point. If the derivative of this function exists at every point, then it defines a new function called the derivative of f and is denoted by f’, df/dx or f'(x). We know that we can perform various operations on numbers. Similarly, we can define algebra for the derivatives of functions, such as sum, difference, product and quotient.

## Algebra or Rules of Derivatives of Functions

The following are the rules called the differentiation rules that represent the algebra of derivatives of functions. These can be applied to solve simple as well as complex problems in calculus and also real life situations.

Suppose f(x) and g(x) are two functions such that their derivatives are defined in a common domain. Then we can define the following rules for the functions f and g.

Sum Rule of Differentiation

The derivative of the sum of two functions is the sum of the derivatives of the functions. This can be expressed as:

$$\large \mathbf{\frac{d}{dx}[f(x)+g(x)]=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)}$$

Difference Rule of Differentiation

The derivative of the difference of two functions is the difference of the derivatives of the functions. This can be expressed as:

$$\large \mathbf{\frac{d}{dx}[f(x)-g(x)]=\frac{d}{dx}f(x)-\frac{d}{dx}g(x)}$$

Product Rule of Differentiation

The derivative of product of two functions f(x) and g(x) is given by the formula:

$$\large \mathbf{\frac{d}{dx}[f(x).g(x)]=f(x)\frac{d}{dx}g(x)+g(x)\frac{d}{dx}f(x)}$$

Also, check: Product rule formula

Quotient Rule of Differentiation

The following formula gives the derivative of the quotient of two functions, provided the denominator is non–zero.

$$\large \mathbf{\frac{d}{dx}\left ( \frac{f(x)}{g(x)} \right )=\frac{g(x).\frac{d}{dx}f(x)-f(x).\frac{d}{dx}g(x)}{[g(x)]^2}}$$

The derivatives of product and quotient of two functions can also be expressed as given below.

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

(uv)′ = uv′ + u′v

Also, (u/v)′ = (u′v – uv′)/v2

These are referred to as a Leibnitz rule for differentiating the product and quotient of functions respectively.

Also, check Leibnitz theorem here.

Go through the solved examples given below to understand how to apply the algebra of derivatives of functions that involve the use of several differentiation formulas.

### Solved Examples

Example 1:

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

Solution:

Given: x + (1/x)

Let f(x) = x 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 [x + (1/x)] = d/dx (x) + d/dx (1/x)

= 1 + (-1/x2)

= 1 – (1/x2)

Example 2:

Find the derivative of sin x – cos x.

Solution:

Given function is: sin x – cos x

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

Using the difference rule of differentiation,

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

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

= cos x – (-sin x)

= cos x + sin x

Example 3:

Find the derivative of (5x3 – 3x + 1)(x + 1).

Solution:

Given: (5x3 – 3x + 1)(x + 1)

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

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

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

= (5x3 – 3x + 1) + (x + 1)(15×2 – 3)

= 5x3 – 3x + 1 + 15x3 – 3x + 15x2 – 3

= 20x3 + 15x2 – 6x – 2

Example 4:

Compute the derivative of f(x) = cot x.

Solution:

Given,

f(x) = cot x

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

Let u(x) = cos 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 (cos x)] sin x – cos x[d/dx (sin x)]} / sin2x

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

= -[sin2x + cos2x]/ sin2x

= -1/sin2x

= -cosec2x