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= ddxx2–ddx(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=3xddxx2–x2ddx3x3x2
Step 2: Apply power rule and solve
ddxx23x=3x(2x2-1)-x23(1)9x2
ddxx23x=6x2–3x29x2
= 6x2–3x29x2
= 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.