# Differentiation Rules

The differentiation rules help us to evaluate the derivatives of some particular functions, instead of using the general method of differentiation. The process of differentiation or obtaining the derivative of a function has the significant property of linearity. This property executes taking the derivative more natural for functions constructed from the primary elementary functions, using the procedures of addition and multiplication by a constant number.

The important rules of differentiation are:

- Power Rule
- Sum and Difference Rule
- Product Rule
- Quotient Rule
- Chain Rule

Let us discuss these rules one by one, with examples. Also, read Differentiation method here at CT’S.

## Power Rule of Derivatives

This is one of the most common rules of derivatives. If x is a variable and is raised to a power n, then the derivative of x raised to the power is represented by:

d/dx(x^{n}) = nx^{n-1}

**Example: Find the derivative of x ^{5}**

Solution: As per the power rule, we know;

d/dx(x^{n}) = nx^{n-1}

Hence, d/dx(x^{5}) = 5x^{5-1} = 5x^{4}

## Sum Rule of Derivatives

If the function is sum or difference of two functions, then the derivative of the functions is the sum or difference of the individual functions, i.e.,

If f(x)=u(x)±v(x), then;

f'(x)=u'(x)±v'(x) |

**Example: f(x) = x+x ^{3}**

Solution: By applying sum rule of derivative here, we have:

f’(x) = u’(x) + v’(x)

Now, differentiating the given function, we get;

f’(x) = d/dx(x+x^{3})

f’(x) = d/dx(x) + d/dx(x^{3})

f’(x) = 1+3x^{2}

## Product Rule of Derivatives

According to the product rule of derivatives, if the function f(x) is the product of two functions u(x) and v(x), then the derivative of the function is given by:

If f(x) = u(x)×v(x), then:

f′(x) = u′(x) × v(x) + u(x) × v′(x) |

**Example: Find the derivative of x ^{2}(x+3).**

Solution: As per the product rule of derivative, we know;

f′(x) = u′(x) × v(x) + u(x) × v′(x)

Here,

u(x) = x^{2 } and v(x) = x+3

Therefore, on differentiating the given function, we get;

f’(x) = d/dx[x^{2}(x+3)]

f’(x) = d/dx(x^{2})(x+3)+x^{2}d/dx(x+3)

f’(x) = 2x(x+3)+x^{2}(1)

f’(x) = 2x^{2}+6x+x^{2}

f’(x) = 3x^{2}+6x

f’(x) = 3x(x+2)

## Quotient Rule of Derivatives

If f(x) is a function, which is equal to ratio of two functions u(x) and v(x) such that;

f(x) = u(x)/v(x)

Then, as per the quotient rule, the derivative of f(x) is given by;

\(\mathbf{f}^{\prime}(\mathbf{x})=\frac{\mathbf{u}^{\prime}(\mathbf{x}) \times \mathbf{v}(\mathbf{x})-\mathbf{u}(\mathbf{x}) \times \mathbf{v}^{\prime}(\mathbf{x})}{(\mathbf{v}(\mathbf{x}))^{2}}\) |

**Example: Differentiate f(x)=(x+2) ^{3}/√x**

Solution: Given,

f(x)=(x+2)^{3}/√x

= (x+2)(x^{2}+4x+4)/√x

= [x^{3}+6x^{2}+12x+8]/x^{1/2}

= x^{-1/2}(x^{3}+6x^{2}+12x+8)

= x^{5/2}+6x^{3/2}+12x^{1/2}+8x^{-½}

Now, differentiating the given equation, we get;

f’(x) = 5/2x^{3/2} + 6(3/2x^{1/2})+12(1/2x^{-1/2})+8(−1/2x^{-3/2})

= 5/2x^{3/2 }+ 9x^{1/2} + 6x^{-½} − 4x^{-3/2}

## Chain Rule of Derivatives

If a function y = f(x) = g(u) and if u = h(x), then the chain rule for differentiation is defined as;

dy/dx = (dy/du) × (du/dx) |

This rule is majorly used in the method of substitution where we can perform differentiation of composite functions.