# What is the basics of Derivative and what are its types?

Derivative are the basic fundamentals of mathematics. It is the rate of change of function with respect to variables. Derivative is used to measure the sensitivity of dependent variable with respect to independent variable. In this post we will learn about the basic concepts of derivative and its types along with examples.

Before starting the types of derivatives, we should familiar with the basic rules of derivatives.

## Rules of derivative

Some basic rules of derivatives are:

 Rules Results Constant rule ddx(C)=0 Power rule ddxxn=nxn-1 Sum rule ddxf(x)+g(x=ddxfx+ddxg(x) Difference rule ddxfx-g(x=ddxfx–ddxg(x) Product rule ddxfx*g(x)=f(x)ddxgx+g(x)ddxf(x) Quotient rule ddxfxg(x)=gxddxfx-fxddxgxgx2 Exponent rule ddxex=ex and ddxef(x)=efxddxf(x) Log rule ddxln x =1x     and      ddxln fx =1f(x)ddxx Constant power rule ddxax=axln⁡(a) and ddxaf(x)=afxln⁡(a)ddxf(x)

## What is the order of derivative?

Before starting the types, we should have some knowledge about the order of derivative.

Order of the differential equation is the order of highest derivative present in the equation. Let us learn about the order through examples.

Example 1

Find the order of  ddx(x3+x) ?

Solution

Since the differential is used in the given problem is the first derivative having power 1

Order of ddxx3+x=1

Example 2

Find the order of  d2dx2(x3+x) ?

Solution

Since the differential is used in the given problem is the second derivative having power 2

Order of d2dx2x3+x=2

Example 3

Find the order of  d3dx3(x3+x) ?

Solution

Since the differential is used in the given problem is the third derivative having power 3

Order of d3dx3x3+x=3

## Types of derivatives

### Explicit differentiation

In explicit differentiation we deal with only one unknown variable. We change the rate of a function with respect to its variable. It is not having any equation if there is an equation in any problem then that problem will be implicit.

Let us learn how to calculate the explicit differentiation with examples. You can also use the online derivative calculator to solve the derivative problems for free.

Example 1

Find the derivative of x2+2x?

Solution

Step 1: write the derivative notation

ddx(x2+2x)

Step 2: Apply sum rule

ddxx2+2x= ddx(x2)+ddx(2x)

Step 2: Apply power rule and solve

ddxx2+2x= 2x2-1+2(1)

= 2x+2

= 2(x+1)

Example 2

Find the derivative of x2-3x?

Solution

Step 1: write the derivative notation

ddx(x2-3x)

Step 2: Apply difference rule

ddxx2-3x= ddxx2ddx(3x)

Step 2: Apply power rule and solve

ddxx2-3x= 2x2-1-3(1)

= 2x-3

Example 3

Find the derivative of x2*3x?

Solution

Step 1: write the derivative notation

ddx(x2*3x)

Step 2: Apply product rule

ddxx2*3x= x2ddx3x+3xddx(x2)

Step 2: Apply power rule and solve

ddxx2*3x= x23(1)+3x(2x2-1)

ddxx2*3x= 3x2+3x(2x)

ddxx2*3x= 3x2+6x2

Example 3

Find the derivative of x2/3x?

Solution

Step 1: write the derivative notation

ddx(x23x)

Step 2: Apply quotient rule

ddxx23x=3xddxx2x2ddx3x3x2

Step 2: Apply power rule and solve

ddxx23x=3x(2x2-1)-x23(1)9x2

ddxx23x=6x23x29x2

= 6x23x29x2

= 3x29x2

= 1/3

### Implicit differentiation

The process of finding dy/dx is known as implicit differentiation. In simple words to the derivative of an equation we use implicit differentiation. In this process we apply the derivative on both sides. In equation we also find the derivative of y with respect to x, the derivative of y with respect to x is dy/dx and the derivative of y2 is 2y*dy/dx.

Example 1

Find the derivative of x3 + y2 = 10

Solution

Step 1: First of all, take derivative on both sides

ddx(x3 + y2) = ddx10

Step 2: Apply the sum rule

ddx(x3) +ddx (y2) = ddx10

Step 3: Now apply the power and constant rule

3x3-1+2ydydx = 0

3x2+2ydydx = 0

2ydydx = -3x2

dydx = -3x2/2y

Example 2

Find the derivative of 2xy + y2 = 2y

Solution

Step 1: First of all, take derivative on both sides

ddx(2xy + y2) = ddx2y

Step 2: Apply the sum rule

ddx(2xy) +ddx (y2) = ddx2y

Step 3: Now apply the product and power rule

2xdydx+2ydxdx+2ydydx = 2dydx

2xdydx+2y+2ydydx = 2dydx

2xdydx+2ydydx-2dydx = -2y

(2x+2y-2)dydx = -2y

dydx = -2y/ (2x+2y-2)

dydx = -y/ (x+y-1)

Example 3

Find the derivative of 4x3 + y2 = 100

Solution

Step 1: First of all, take derivative on both sides

ddx(4x3 + y2) = ddx100

Step 2: Apply the sum rule

ddx(4x3) + ddx (y2) = ddx100

Step 3: Now apply the power and constant rule

4*3x3-1+2ydydx = 0

12x2+2ydydx = 0

2ydydx = -12x2

dydx = -12x2/2y

dydx = -6x2/y

Example 4

Find the derivative of 2xy + y2 = 2y2

Solution

Step 1: First of all, take derivative on both sides

ddx(2xy + y2) = ddx2y2

Step 2: Apply the sum rule

ddx(2xy) + ddx (y2) = ddx2y2

Step 3: Now apply the product and power rule

2xdydx+2ydxdx+2ydydx = 4ydydx

2xdydx+2y+2ydydx = 4ydydx

2xdydx+2ydydx-4ydydx = -2y

(2x+2y-4y)dydx = -2y

(2x-2y)dydx = -2y

dydx = -2y/ (2x-2y)

dydx = -y/ (x-y)

### Partial differentiation

When a function has several variables, we use partial differentiation in which all the variables vary. It is mostly used in differential geometry and vector calculus.

Partial derivatives are denoted in many ways such as

f’x, Dxf, xf ∂xf or ∂f∂x

Let us learn about partial derivatives through some examples.

Example 1

Find the partial derivative of 2xy + y2 with respect to x?

Solution

Step 1: First of all, take partial derivative

∂x(2xy + y2

Step 2: Apply the partial derivative only on the function having x remaining all consider as constant.

∂x(2xy) +∂x (y2

2y∂x(x) +∂x (y2

2y(1)+ 0

∂x(2xy + y2) = 2y

Example 2

Find the partial derivative of 2xy + y2 with respect to y?

Solution

Step 1: First of all, take partial derivative

∂y(2xy + y2

Step 2: Apply the partial derivative only on the function having y remaining all consider as constant.

∂y(2xy) +∂y (y2

2x∂y(y) +∂y (y2

2x(1)+ 2y

∂y(2xy + y2) = 2x + 2y

Example 3

Find the partial derivative of 2xy + y2 with respect to xy?

Solution

Step 1: First of all, take partial derivative

∂xy(2xy + y2

Apply the partial derivative only on the function having xy remaining all consider as constant.

Step 2: First take the partial derivative with respect to x

∂x(2xy) +∂x (y2

2y∂x(x) +∂x (y2

2y(1)+ 0

∂x(2xy + y2) = 2y

Step 3: Now take the partial derivative of above result with respect to y

∂y(2y) = 2

Step 4: write the result along with given problem

∂xy(2xy + y2) = 2

## Summary

Now you have come to know that types of derivatives are not difficult or confusing. We have learnt explicit, implicit and partial differentiation along with lots of examples if you grabbed all the basic concepts about these topics, you will master them in few days.

Categories CG