Differentiation Lecture -1| differentiation class 11 math concept | Mathematically inclined | Constant function rule | Power rule | Basic rule of derivatives | Derivatives of functions | Product rule | Quotient rule

What are derivatives? What are the Basic Rules of derivatives? What are the Derivatives of functions? What are the Constant function rule, Power Rule, product rule, and Quotient rule of derivatives? 

If you are a maths student studying derivatives and want to learn mathematics from the beginning. This lecture is very important for you. In this lecture, I discussed the basics of derivatives in detail and gave the answers mentioned above. This lecture helps to differentiate of class 11, Derivatives for class 12, and derivatives beginners.

What Are Derivatives?

The rate of change of a function with respect to a variable.

In simple words, If two quantities are interrelated to each other. One quantity is dependent upon the other quantity. If there is a change in the first quantity. So because of the change in the first quantity, The amount of change or rate of change in the second quantity is called Derivatives or Differentiation.

As an example:

A person starts their journey from a point "A" after sometimes reaching point "B". The distance covered by the person from A to B is called average distance because this distance gives the information about total distance in total time. 

But if I ask about any random point C that tells me what distance has been covered by you? and what was the time? when you reached any point C. This point C gives you sharp information about a specific place. This is called the rate of change in a quantity with respect to other quantities, as distance changes with respect to time which is an instantaneous change. This rate of change is called Derivatives.

  

Let's take another example:

Consider a circle C1 whose radius is "r". As we know the area of a circle will be  "A= πr^2". 

If I increase the radius of the circle "h". Now the new radius of the circle will be "r+h". So, the area of the circle will be "A=π (r+h)². 

We can see here, that the area of the circle is dependent on the radius. As the radius changes instantaneously h, then the area of the circle converts into π(r+h)² from πr^2. This instantaneous change in the area of a circle with 0respect to radius is called derivative.

Basic Rule Of Derivatives:

The basic rule of derivatives is applicable to any function which has a power of more than 1.

Thus rule can be defined as, considering a given function is

F(x) = x^n 

Where f(x) is the representation of a given function. X can be any function like Trigonometric, exponential, polynomial, etc, and on is a constant value.

When you apply derivatives operation, this value will be converted as

F'(x) = nx^n-1

It means that when you apply a derivative of a function. Power will go in multiplication with function and 1 will be minus from the given power of the function.

 

As an example

Apply derivatives on x^5. 

For solution:

Consider giving function is equal to "y"

y = x⁵

Apply d/dx on both sides 

dy/dx = d/dx (x⁵)

Now power will go in multiplication with the function and 1 will be minus from power.

dy/dx = 5x⁴‐¹

Simplifying 

dy/dx = 5x⁴

Example:

Apply derivatives on 1/x

If the function is a fraction value.

For solution:

First, you should change the friction value into a normal value.

In 1/x "x" has a power of "+1" and exists in the denominator. Now, first, you should move x from the denominator to the numerator. For this purpose power of x will be transformed from "+1" to "-1" and Consider the whole function is equal to y

Y = x–¹

Now apply the basic rule of derivatives 

Power will go In multiplication and 1 will be minus from power.

dy/dx = (-1) x–¹–¹

dy/dx = -1x–²

Now make the power of x positive by moving x from numerator to denominator.

dy/dx = -1/x²

Example:

Apply derivatives on √X

If given function power's in a fraction 

For solution:

Consider the function is equal to y and change root into power value.

y=x½

Now apply derivatives on both sides. Power will go in the given function and 1 will be reduced from power.

dy/dx = ½x^½-¹

dy/dx = ½ x‐½

dy/dx =1/2x½

Quick Rule Of Derivatives (heading)

1- d/dx {f(x) + g(x)} = d/dx f(x) + d/dx g(x)

If two or more functions are given addition or Subtraction for derivative. Derivatives will be applied to each function individually.

Example

Apply derivatives on x² + lnx

Solution:

For x² power will go in multiplication with function and 1 will be reduced from power. For lnx, it will change into 1/x.

dy/dx = (2x) + 1/x

dy/dx = 2x + 1/x

2- d/dx {k f(x)} = k d/dx f(x)

If a given function is in multiplication with any constant value k then, you should leave the constant alone and Apply derivatives on the given function.

Example

Apply derivatives on 2x⁶.

Solution:

Consider function is equal to y:

y = 2x⁶

Apply derivatives on both sides:

dy/dx = d/dx(2 x⁶)

dy/dx = 2 d/dx x⁶ 

dy/dx = 2(6x⁶‐¹)

dy/dx = 12x⁵

3- d/dx {f(x) × g(x)} = f(x)d/dx g(x) + g(x) d/dx f(x)}

If two functions are given in multiplication. For taking derivatives leave the first function as it is and take the derivative of the 2nd function then sign mark as plus then leave the second function as it is and take derivatives of the first function then simplify.

Example: Apply derivatives on xsinx.

Solution: 

Consider function is equal to y

y = xsinx

Apply derivatives on both sides 

dy/dx = x d/dx (sinx) + sinx d/dx x

You can see in the above equation the first function is "x" and the second function is "sinx". Leave x as it is and apply derivative of sinx then mark + sign then leave sinx as it is and apply derivatives on x.

dy/dx = x cosx + sinx (1)

dy/dx = xcosx + sinx

4- d/dx f(x)/g(x) = g(x) f'(x) - f(x) g'(x)/ g²(x)

If two functions are given in division. For taking derivatives leave denominator function g(x) as it is and taking derivatives if numerator function f(x) then mark sign minus then leave f(x) as it is and take derivatives of denominator function g(x). In the denominator write the square of the denominator function.

Example: Apply derivatives on x/lnx

Solution:

Consider function is equal to y 

y = x/lnx

Apply derivatives on both sides

dy/dx = lnx d/dx (x) - x d/dx (lnx) /(lnx)²

In the equation for applying derivatives leave the denominator function same and apply derivatives on numerator function x then mark as a minus sign then apply derivatives on denominator function sinx and leave x same. Square of lnx writes In denominator.

dy/dx = lnx (1) - x(1/x) / (lnx)²

dy/dx = lnx - 1 /(lnx)²

Mix Question No.1:

Apply derivatives on y= lnx + sinx - x²

Solution:

Consider function is equal to y

y = lnx + sinx - 3x²

Apply derivatives on both sides 

dy/dx = 1/x + cosx - 3(2x)

dy/dx = 1/x + cosx - 6x.

Mix Question No.2:

Apply derivatives on xsinx + cos²x

Solution:

Consider function is equal to y 

y = xsinx + cos²x

Apply derivatives 

dy/dx = (x d/dx sinx + sinx d/dx x) + 2cosx d/dx cosx

In the above equation (xsinx) will be open with quick Rule number 3 which is a product of two functions and cos²x will be open with power rule

dy/dx = (x cosx + sinx) + 2cosx (- sinx)

dy/dx = xcosx + sinx -2cosxsinx

For Video Lecture:

Algebra1 Lesson1: ttps://a-maths1.blogspot.com/2022/02/equation-solving1.html

Algebra1 Lesson2: https://a-maths1.blogspot.com/2022/03/step-two-equation.html