concept on derivatives | what is the derivatives ? | rules of derivatives | mathematics (+2 level)

-

 One of the difficult but interesting and very useful topics in mathematics is derivatives.

Derivatives give you an insight into the change of a function with respect to its variables.

For example, change of position of a car with respect to time. 

The market value of any products with respect to the quantity of production and production cost.

The demand for daily used goods with respect to the number of people in the locality.

All these aspects represent a kind of derivatives. 

For derivative, one needs to calculate the limit of the ratio of the increment of a function (`\Deltaf`) to the corresponding increment of the independent variable (`\Deltax`) of the function when `Deltax` tends to zero.    

Mathematically,

`\lim_(\Deltax->0) \frac{\Deltaf}{\Deltax} = \frac{df(x)}{dx} ......(1)`

The method of calculating the limit is an operation that is called as differentiation of the function with respect to its variables. 

We denote y = f(x), so

`\frac{df(x)}{dx} = \frac{dy}{dx}`

The result obtained from this operation is also called the differential coefficient of the function. 

The differential coefficient for f(x) in the interval (a, b) is also expressed as,

`\lim_(\Deltax->0) \frac{f(x + \Deltax) - f(x)}{\Deltax}`

This is called first principle(limit theorem) of derivatives.

Some of the symbols used to denote derivative of y = f(x) with respect to x are

`f\prime(x), y\prime, \frac{df(x)}{dx}, \frac{dy}{dx}`

When you go on differentiating a function, power or degree or index (for example, 3 in `x^3`) of variables involved in the function goes on decreasing.



For example:-

Let's calculate derivatives of x, `x^2` and `x^3`.

1. y = f(x) = x

Let `\Deltax` be a small increment in x and `\Deltay` be a small increment in y, then

`y + \Deltay = x + \Deltax`

or, `\Deltay = x + \Deltax - y`

or, `\Deltay = x + \Deltax - x`

or, `\Deltay = \Deltax`

or, `\frac{\Deltay}{\Deltax} = 1` 

`\therefore` `\frac{dy}{dx} = \lim_(\Deltax->0) \frac{\Deltay}{\Deltax} = 1`

Thus, the derivative of y = f(x) = x is 1 i.e. `x^o` which shows that power of x is reduced from 1 to 0.// 

2. `y = f(x) = x^2`

Let's `\Deltax` and `\Deltay` be small increments in respective x and y. Then

`y + \Deltay = (x + \Deltax)^2`

or, `\Deltay = x^2 + 2x\Deltax + \Deltax^2 - y`

or, `\Deltay = x^2 + 2x\Deltax + \Deltax^2 - x^2`

or, `\Deltay = \Deltax (2x + \Deltax)`

or, `\frac{\Deltay}{\Deltax} = 2x + \Deltax`

or, `\frac{dy}{dx} = \lim_(\Deltax->0) \frac{\Deltay}{\Delta} = 2x + 0`

Because `\Deltax` is constant and derivative of constant quantity is zero.

`\therefore` `\frac{dx^2}{dx} = 2x`.

Here you see that power of x is reduced from 2 to 1. Also, the initial value of power goe on multiplying the result after the differentiation.//

Following the same process for power 3, you get

`\frac{dx^3}{dx} = 3x^2`. //

Derivative of  `x^n` first principle(limit theorem):-

Let's `y = x^n ......(1)` 

Consider `\Deltax` and `\Deltay` are respective small increment in x and y, then 

`y + \Deltay = (x + \Deltax)^n ..... (2)`

Subtracting (1) from (2),

`\Deltay = (x + \Deltax)^n - x^n`

or, `\frac{\Deltay}{\Deltax} = \frac{(x + \Deltax)^n - x^n}{\Deltax}`

Now, 

`\frac{dy}{dx} = \lim_(\Deltax->0) \frac{(x + \Deltax)^n - x^n}{\Deltax}`

or, `\frac{dy}{dx} = \lim_(\Deltax->0) \frac{(x + \Deltax)^n - x^n}{(x + \Deltax) - x}`

or, `\frac{dy}{dx} = \lim_(x+\Deltax ->x)\frac{(x+\deltax)^n-x^n}{(x+\Deltax)-x}`

Using the result, `\lim_(x->a) \frac{x^n-a^n}{x-a} = na^{n-1}`, we get

`\frac{dy}{dx} = nx^{n-1}`.

`\therefore` `\frac{dx^n}{dx} = nx^{n-1}`. //

Derivative of `(ax+b)^n` by limit theorem:-

let `y = (ax+b)^n .......(1)` 

Also, `y+\Deltay = (a(x+\Deltax) + b )^n .......(2)`

Subtract (1) from (2),

`\Deltay = [a(x+\Deltax) +b]^n - (ax+b)^n`

or, `\frac{Deltay}{\Deltax} = \frac{(ax+a\Deltax +b)^n - (ax+b)^n }{\Deltax}`

Now,

`\frac{dy}{dx} = \lim_(\Deltax->0) \frac{(ax+a\Deltax +b)^n - (ax+b)^n}{\Deltax}`

or, `\frac{dy}{dx}=\lim_(\Deltax->0)\frac{(ax+a\Deltax +b)^n-(ax+b)^n}{a\Deltax}\cdota`

or, `\frac{dy}{dx}=\lim_(ax+a\Deltax+b->ax+b)\frac{(ax+a\Delta+b)^n-(ax+b)^n}{(ax+a\Deltax+b)-(ax+b)}`

or, `\frac{dy}{dx}=n(ax+b)^{n-1} a`

`\therefore` `\frac{d(ax+b)^n}{dx} = na(ax+b)^{n-1}` //

Derivative of `y =e^x` by limit theorem: 

Here, `y =e^x .......(1)`

Also, `y + \Deltay = e^{x + \Deltax} .......(2) `

Subtract (1) from (2) you get,

`\Deltay = e^{x + \Deltax} - e^x`

or, `\Deltay = e^x \cdot e^{\Deltax} -e^x`

or, `\Deltay = e^x(e^{\Deltax} - 1)`

or, `\frac{\Deltay}{\Deltax} = \frac{e^x(e^{\Deltax}-1)}{\Deltax}`

Now,

`\frac{dy}{dx} = \lim_(\Deltax->0)\frac{\Deltay}{\Deltax} = \lim_(\Deltax->0)\frac{e^x(e^{\Deltax}-1)}{\Deltax}`

Here `e^x` can be taken out from limit.

`\frac{dy}{dx}=e^x \lim_(\Deltax->0)\frac{e^{\Deltax}-1}{\Deltax}`

Using limit theorem, `\lim_(x->0)\frac{e^x  1}{x} = 1`,

`\frac{dy}{dx} = e^x \cdot 1`

`\therefore` `\frac{de^x}{dx} = e^x` 

You can also prove that

`\frac{de^{ax}}{dx} = ae^{ax}` //

Derivative of `y = log_ax`:

Here, `y = log_ax  = log_a(x)` 

'a' in the above equation is called the base of a logarithmic function. 'a' can be equal to 10, e, or any numbers. 

Using the property, `log_a(x) = log_ae \cdot log_e x`

This is how you can change base of log function from a to e.  

`y = log_ae log_ex `

You can write `log_ex` simply as logx. See carefully when e of `log_ae` cancels e of `log_ex` gives you `log_ax` again.

`y = log_ae \cdot logx ....(1)`

Let `\Deltax` and `\Deltay` be small increments in x and y respectively. Then

`y+\Deltay =log_ae log_a(x+\Deltax) .....(2)` 

Subtract (1) from (2),

`\Deltay = log_ae(log_a(x+\Deltax)-log_ax)`

Divide both sides by `\Deltax`,

`\frac{\Deltay}{\Deltax} = log_ae \frac{log_a(x+\Deltax)-log_ax}{\Deltax}`

Now,

 `\frac{dy}{dx} = \lim_(\Deltax->0)\frac{\Deltay}{\Deltax} =log_ae  \lim_(\Deltax->0)\frac{log_a(x+\Deltax)-log_ax}{\Deltax}`

Using property of log function, `log_am - log_an = log_a(\frac{m}{n})`,

`\frac{dy}{dx} = log_ae\lim_(\Deltax->0)\frac{log_a(\frac{x+\Deltax}{x})}{\Deltax}`

or, `\frac{dy}{dx} =log_ae \lim_(\Deltax->0)\frac{log_a(1 + \frac{\Deltax}{x})}{\Deltax}`

or, `\frac{dy}{dx} =log_ae \lim_(\Deltax->0)\frac{log_a(1 + \frac{\Deltax}{x})}{\frac{\Deltax}{x}x}`

Taking `\frac{1}{x}` out of lim and using the result, `\lim_(x->0)\frac{log_a(1+x)}{x} =1`,

`\frac{dy}{dx} = log_ae \frac{1}{x}\cdot 1`

`\therefore` `\frac{dlog_ax}{dx} = log_ae \frac{1}{x}` //

Here you also learnt that `\frac{d log}{dx} = \frac{1}{x}`//


To learn derivatives of trigonometric functions: click here.


Rules of finding Derivatives:

Till now, you learn to find derivatives of only a single function f(x) that contains only one variable (x). In this section, you will learn techniques to find derivatives of two or more than two functions.

Rule 1: The Sum/difference rule

Suppose, u(x), v(x), w(x) are functions of a variable x, so that

`y = f(x) = u(x) \pm v(x) \pm w(x)` then derivative of f(x) is

`\frac{d f(x)}{dx} = frac{d u(x)}{dx} \pm \frac{d v(x)}{dx} \pm \frac{d w(x)}{dx}`

Rule 2: The Product rule

Suppose u(x), v(x) and w(x) are functions of x, so that

`y= f(x) = u(x) \cdot v(x) \cdot w(x)` 

f(x) is the product of u(x), v(x) and w(x).Then derivative of f(x) with respect to x is,

`\frac{df(x)}{dx}= v(x) w(x) \frac{du(x)}{dx} + u(x) w(x) \frac{dv(x)}{dx} + u(x) v(x) \frac{dw(x)}{dx}`

Rule 3: The Power rule

Suppose `u(x)^n` function of x with power n so tha

`y = f(x) = u(x)^n`

This means y= f(x) is the product of u(x) multiplied by itself n times. Then derivatives of f(x) with respect to x is, 

 `\frac{d f(x)}{dx} = n u(x)^{n-1} \frac{d u(x)}{dx}`

Rule 4: The Chain rule

If f(u) is function of u and u(t)  is function of t, again t is function of x, so that
`y  = f(x) = u(t)`

Then derivative of f(x) with respect to x is given as

`\frac{d f}{dx} = \frac{d u(t)}{dx}`

or, `\frac{d f}{dx} = \frac{d u(t)}{dt} \cdot \frac{d t(x)}{dx}`

This is called chain rule.

Rule 5: The Quotient rule

Suppose f(x) is composed of u(x) and v(x) such tha

`f(x) = \frac{u(x)}{v(x)}` then derivative of f(x) with respect to x is given as

`\frac{d f(x)}{dx} = \frac{\frac{u(x)}{v(x)}}{dx}`

or, `\frac{df(x)}{dx} = \frac{v(x) \frac{du(x)}{dx} - u(x)\frac{dv(x)}{dx}}{v(x)^2}`

This is called quotient rule, because function is in quotient form.

To learn these rules with examples, click here.

Have a glance at each rule carefully to understand the differences and applications of these rules.

Comments

Post a Comment

These posts may also be useful for you

top 3 websites for downloading research papers for free

-
  This post provides you valid links of top 3 websites from where you can download research papers published on high ranked journals for free. Just copy the Title or DOI of the papers, then go to either of these websites and paste on search bar and then hit search button.  1. Sci -hub  2. Library genesis 3.Core Download the file to get the links of these websites. Get File Doing research works is not only learning, it's innovation and interpretation of field of interest.  Watch video 👇👇 Watch Now Don't forget to Follow this blog.  Follow this Blog  
Read More »

plot graph in gnuplot from csv and data file | knowledge of physics

-
 If you are using LINUX based OS, for example Ubuntu, you must be familiar with plotting data in gnuplot saved in a file with .dat extension. It is very simple technique.  To plot data in .dat format, you simply go to gnuplot and run following command, gnuplot> plot 'file_name.dat' using 1:2  This command gives you simple plot of your data saved in  file_name.dat. You can further modify this plot by adding title, axes names, legends and colors of plot and width etc. Here 1:2 means first column of data is along x-axis and second column along y-axis. But what about if you have data in a file of  .csv format ? Let's discuss. CSV means coma separated variables. This means each column of the data in the .csv file is separated by coma (,), and gnuplot does not accept this format directly just as you do for the .dat file.  For gnuplot to read data from .csv file, you should tell it that data you are applying are in coma separated format. So, write code as below...
Read More »

projectile motion in fortran | relation of angle of projection and horizontal range

-
Image
 Have you ever tried to visualize projectile motion in fortran program? This post will help you to generate data and plot the graph of projectile motion.   Above figure shows you the angle of projection(°) versus horizontal range (meter) plot, where the speed of projection is 30 m/s. 👉 Click on link for NEB grade 12  physics numerical problem-solutions:-  LINK Let's start with some basic things: What is projectile and projectile motion ? Projectile:  A projectile is an object thrown to the sky with certain velocity against the earth's gravity. The motion that the projectile possesses during it's journey until it returns to the same surface, from where it was thrown, is called the projectile motion. In above figure, the projectile starts it's journey from point o and follows the curved path to reach another point on the same surface.  Suppose, `\theta` = angle of projection `\therefore` Angle of projection is the angle between the horizontal surface of ...
Read More »

NEB physics exam numerical problems with solutions for grade 12 students

-
Image
 This post provides you the numerical problems and their solutions of physics based on NEB(National Examination Board) syllabus grade 12. Solutions of Previously Asked Numerical Problems in NEB exams. Q.1. A coil consisting of 100 circular loops with radius 0.60 cm carries a current of 5A. At what distance from the center, along the axis, the magnetic field has magnitude as `\frac{1}{8}` great as it is at the center? Solution:- Given, number of circular loops = N = 100 radius of the loop, `r = 0.60cm = 6.0 \times 10^-3` m. current, I = 5 A Magnetic field at the center of loop = B (suppose) magnetic field at distance x from the center = `\frac{1}{8}B`   To find, x = ? The magnetic field at distance x from center of the coil is given as, `B = \frac{\mu_o I N r^2}{2(x^2  + r^2)^(\frac{3}{2})}`. By given in question, magnetic field at x = `\frac{1}{8} \times` magnetic field at x = 0.       or, `\frac{\mu_o I N r^2}{2(x^2 + r^2)^(\frac{3}{2})} =\frac{1...
Read More »

how to plot 3D and parametric graphs in gnuplot | three dimensional plot in gnuplot software

-
Image
Although being an open-source plotting software, gnuplot gives you almost all feasibilities to plot 2D, 3D, parametric plots and even the gif as well as animations.  Here you can learn to plot three dimensional graphs in gnuplot. Additionally, you also get the idea of parametric plots in gnuplot software. 1. 3D graphs in gnuplot To plot 3D graphs you just need to replace the command 'plot' for 2D graphs by 'splot' for 3D graphs.  Here, splot means surface plot.  Plotting graphs in 3D view allows you to observe it from all possible sides of the graph.  You rotate it upward, downward , left-side as well as to right-side.  Example: gnuplot >  splot sin(x) Then press ENTER key. You get the output as it looks as picture below: Fig. 1 3D graph of sin(x). 2. parametric plots Parametric plots can be created both in 2D and 3D.  For 2D graph, the dummy variable is t, whereas for 3D plot, they are u and v. Example: Image of parametric plot is shown below: Fig....
Read More »

how to create GIF animation in gnuplot | animation using gnuplot software

-
Image
 GNUPLOT software is not only 2D and/or 3D plotting tool, but it also provides flexibility to create animation and GIF animation plots.  GIF means Graphics Interchange Format. It is a repeated display of images from same of different sides. It helps you to view the image clearly so that you don't need to pay much attention to understand it.  This post provides you tools and techniques to create GIF animation in gnuplot. Codes necessary for creating the GIF are given in links. Please check it as well. GIF can also be both 2D and 3D.  You must use 'plot' command to create 2D GIF and 'splot' command to create 3D GIF. 👉 To learn to create plots in gnuplot: Click Here   👉 To learn gnuplot basics for beginners: Click Here Example of 2D GIF: GNUPLOT code for creating above GIF, 👇 gnuplot > set title 'gif image of sine and cosine functions' gnuplot > set xlabel 'x' gnuplot > set ylabel 'sine and cosine' gnuplot > f1(x,t) = sin(x-t) gnu...
Read More »

NEB board exam maths solved problems with proper solutions - available for free | short answered mathematics problems solved group A

-
Image
 This post provides you the problems with solutions of NEB board exam of grade 11 mathematics 2076. Attempt All Questions. Group A  `5 \times 3 \times 2 = 30` Get full question paper from Here . 1. a. Find the truth value of the biconditional statements of p and q where p represents '2 is an even number' and q represents '4 is an even number'. Solution:- Biconditional statements imply if one statement is true then another is also true. if one is false, then another statement is also false. So, one statement supports the other in case of biconditional statement.  Here, p: 2 is an even number.          q: 4 is an even number. 4 is the square of 2 and 2 is the square root of 4. 2 is even, then 4 is also even. So, both p and q represent true value (T). p q p ⇔ q T T T Thus truth value of p and q is T, i.e. 2 is an even number implies 4 is also an...
Read More »

NEB grade 11 and 12 maths exam MCQ solved problems | NEB mathematics MCQ solved problems

-
Image
 NEB  Mathematics grade 11 and 12 specification grid solutions Short Solution Problems(Group A) Rewrite the correct option in your answer sheet. 1. Which of the following is a statement? a) The fishes are beautiful. b) Study Mathematics c) x is the capital of country y. d) Water is essential for health. Correct option: (d) It because any sentence is considered as a statement if it is either true or false but not both. Among given options, (a), (b) and (c)  do not declare truth or falsity. But option (d) declares the truth that water is important for health. 2. The  value of: `\sqrt{-16} \times \sqrt{-25}` is,  (a) -20 (b) -20i  (c) 20i  (d)20 Correct option: (a) Here,  `\sqrt{-16} \times \sqrt{-25}` = `\sqrt{-1 \times 16} \times \sqrt{-1 \times 25}`  But `i^2 = -1`, i = complex number. = `\sqrt{i^2 \times 4^2} \times \sqrt{i^2 \times 5^2}` = ` i \times 4 \times i \times 5` = `i^2 \times 20` = -20. // Click here for NEB grade 11 maths short so...
Read More »