Please, someone explain to me in simple language whats a derivative and how to find one of a fuction?

Question

give some examples.and use equations

Answer 1

In math, a derivative is the rate of change.

For example, the equation y=2x has a constant rate of change of 2. For every 2 "x" (or run, as it sometimes refered), there is 1 "y" (or rise).

Similarly, in parabolic equations, the change is described by the derivative. If the equation was y = x^2 (X squared), then the derivative, or y' (y-prime) would be 2x. That means that the change would be relative to the location of "x" on the curve. At x=1, the change is 2. At x=10, the change is 20.

There is no easy answer as to how to find the derivative for a function, because that is the intricacy of calculus. "Simple" (F(x)=ax^n, where n is a whole number) functions are relatively easy, but trig functions, log functions, and others take some work to understand.

Answer 2

A derivative is a function representing the slope of the first.

You remember how to find the slope of a line right? Now, if you have most any function you can find the slope at a particular point. Lets do a thought exercise. If you plot on a piece of paper the function y=x^2 very carefully from x=-3 to x=3 you can set a rule on the piece of paper so that it is just barely touching the curve you have drawn. If you then draw a line along the edge of the ruler you can measure the slope of the line you just drew.

Trying this for a few points gives:
x x^2 slope
-3 9 -6
-2 4 -4
-1 1 -2
0 0 0
1 1 2
2 4 4
3 9 6

If you didn't get those exact answers for the slope it's ok, it just means that you didn't line up the ruler or draw your y=x^2 perfectly. Trust me that these are the values you should have gotten if everything went just right.

Next, let's try plotting those slopes:
x y
-3 -6
-2 -4
-1 -2
0 0
1 2
2 4
3 6

You should be able to draw a new function connecting all of those points. It is y=2x. This new function is the derivative; it is a representation of the slope of the first function at every point. Therefore, we say that 2x is the derivative of x^2.

That is the basic idea. The rest of differential calculus is just trying to calculate derivatives precisely.

Answer 3

In simple language.

Imagine you took two roller coaster rides.

The first one is on a completely straight inclination at a fixed slope.

The second one is more exciting, it's like the usual curves with ups and downs.

Now, I ask you, what was your experience taking the first ride compared with the second?

And your Answers:

1. The first one was quite boring, there was no change in the slope (gradient), it just went down straight.

2. The second one was full of excitment, the slope (gradient) was changing all the time. Sometimes, it slows down and at other times, it was very steep and fast.

Now, the derivative of a function is understanding how much changes there are in a slope. If there are no changes, the derivative is equals to zero (straight line).

Other slopes, change from gentle to steep and then steep to gentle and so on.

Hope you have an enjoyable ride on your fascinating roller coaster journey of Calculus.

Answer 4

it is true that a derivative is the rate of change of a function with respect to a variable. In the case y = f(x), the derivative of f(x) is the change in y for a change in x. Unless the function is a straight line, this will vary depending on the value of x, so we consider the change in y for only a very small change in x (we wil call this change in x "dx"). Therefore the derivative is defined as

lim[dx -> 0] of {[f(x+dx) - f(x)] / dx}.

It is easier to understand just by applying the above formula.

Other answerers have said that the derivative of x^2 is 2x; but did not say why that is. We can derive that by applying the above formula:

f(x+dx) = (x+dx)^2 and f(x) = x^2

the derivative is then lim(dx->0) {[(x+dx)^2 - x^2] / dx}

Square the binomial x+dx to get x^2 + 2xdx + x^2

Then the derivative is lim(dx->0) [(x^2 + 2xdx +dx^2 - x^2) / dx]

= lim(d ->0) [(2xdx *dx^2) / dx] = lim(dx->0) [2x + dx] = 2x

Answer 5

A 'derivative' is a number that express how 'quickly' a function is changing at any point on its' graph. Geometrically, it's the slope of a line tangent to the function. The definition for the derivative of a function f(x) is
f'(x) = lim δ ->0 of (f(x+δ)-f(x))/δ

Suppose f(x) = 3x, then
f'(x)=lim δ->0 of (3(x+δ)-3x)/δ = (3x+3δ-3x)/δ = 3δ/δ = 3

You can go through the same thing and discover that
f(x) = k (k a constant) f'(x) = 0
f(x) = x² f'(x) = 2x

and so on. All of these are in your book.

**Study** them. That doesn't mean glance at them and say, "Kewl". That means set down and work throught the example problems and then work through the practice problems.

It also doesn't hurt if you take notes in class ☺


Doug

Answer 6

What you ask is one half of one semester course also known as Calculus I. Do yourself a favor and get a calculus book and read about it. No one is going to learn derivatives from a one page answer.

Answer 7

Do you wish me to bring you a cup of tea while I'm at it?
Where is your Calculus text? Did you feed it to the dog AGAIN?

From any calc text:

The derivative of a function f(x) with respect to x is the limit of (f(x)-f(x0))/(x-x0) as x approaches x0 (x-->x0).

Getting a "feel" for this would take up too much space, so I omit it.

Two easy functions:

y=x^2
dy/dx = lim[(x^2-x0^2)/(x-x0)] as x-->0
Note that x approaches x0 but you never quite let it get there because division by 0 is undefined.
=lim[(x+x0)(x-x0)/(x-x0)], x-->x0
= lim[x+x0], x-->x0 = 2x0 or 2x

dy/dx=d(x^2)/dx = 2x

more generally,

let y=ax^2+bx+c
dy/dx = lim[a(x^2-x0^2)+b(x-x0)+c-c]/(x-x0), x-->x0
=lim[a(x+x0)+b],x-->0
dy/dx = 2ax+b

The more complex the function, the more fun you can have finding the derivative!

Answer 8

Calculus? (shiver) You are going to have to offer more than 10 points for me to go down that road again

Answer 9

Look here: