Science And Technology » The Derivative. http://oscience.info The ultimate resource for Science and Technology Tue, 08 May 2012 16:59:04 +0000 en hourly 1 http://wordpress.org/?v=3.3.2 Derivatives of Logarithmic and Exponential functions. http://oscience.info/mathematics/derivatives-of-logarithmic-and-exponential-functions/ http://oscience.info/mathematics/derivatives-of-logarithmic-and-exponential-functions/#comments Fri, 07 Jan 2011 16:02:37 +0000 subash http://oscience.info/?p=524 Exponential functions are the function which are defined in the form of:

f(x)=ax , where a is a constant and “x” is a variable.

The function “f(x) = ax“ is called an exponential function in base “a”.

The logarithmic functions are the inverse function of exponential function.

Or , if ” y = f(x) = ax ” then , x=f-1(y)  is called the logarithmic function and is denoted by

y= log a x , which is called the logarithmic function in base “a”.

And , The natural exponential is the exponential function when the function is in the base of a  constant “e”.

or, f(x) = ex is called natural exponential function , where the constant  “e” is defined by :

e = \displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n

And the value of “e” is , an irrational number which value is approx.” 2.71828182845904523536 ”

The inverse of natural exponential function is called natural logarithmic function , which is defined by:

y = log e x

for ease the natural logarithmic function is also written by excluding the base “e” ( log x) and
also by replacing log with ln ( ln x).

Derivative of Natural Logarithmic function:

By the definition of derivative:

\frac{d}{dx} (log x) = \displaystyle\lim_{\Delta x\to o} \dfrac{log(x+\Delta x) -log x}{\Delta x}

Now using the properties of logarithms:

= \displaystyle\lim_{\Delta x\to o} \frac{1}{\Delta x} \times log \frac{x+\Delta x}{x}

= \displaystyle\lim_{\Delta x\to o} log \left(\frac{x+\Delta x}{x}\right)^{\frac{1}{\Delta x}}

Now, If we replace \frac{1}{\Delta x} by “v” or , v= \frac{1}{\Delta x}

Then, as \Delta x \to 0 , z \to \infty

\frac{d}{dx} log x = \displaystyle\lim_{z\to\infty} log \left(1+\frac{1}{z}\right)^{\frac{z}{x}}

= \frac{1}{x} \displaystyle\lim_{z\to\infty} log \left( 1+\frac{1}{z}\right)^z

Now , as e = \displaystyle\lim_{z\to\infty} \left( 1+\frac{1}{z}\right)^z

we can write above equation as:

= \frac{1}{x} log e

And as , natural logarithm of “e” is 1.

= \frac{1}{x}

Thus:

The derivative natural logarithmic function is:

\frac{d}{dx} log x = \frac{1}{x}

Derivative of Logarithmic function:

If “y= log a x” is a logarithmic function in base “a”.

We can also re-write the function as: y = log_a e \times log x or , y = log a e . log x, by using the properties of logarithms .

And as “a” and “e” both are constants “log a e” will also be a constant so while differentiating we can take the “log a e” out of the differentiation as “log a e” is a constant.

So,

\frac{d}{dx} log_a x = log_a e \times \frac{d}{dx} ( log x)

And as we have already derived the derivative of natural logarithms, we can differentiate the natural logarithm in the equation which give us:

\frac{d}{dx} log_a x = \frac{log_a e}{x}

Derivative of Natural Exponential function:

We know , y =ex is the natural exponential function.

We can also write it’s inverse function as: x = log y

Now let’s differentiate both side of “x = log y” with respect to “x”:

\frac{d}{dx} x = \frac{d}{dx} log y

Now using the chain rule:

or, 1 = \frac{1}{y} \times \frac{d}{dx} y

We cam re-arrange above equation as:

\frac{d}{dx} y = y

Thus we found the derivative of natural exponential function which is:

\frac{d}{dx} y = y or , \frac{d}{dx} e^x = e^x

Derivative of Exponential function:

If y = ax is a exponential function in base “a” .

As we know a= “e log a “ We can rewrite the function y = ax as:

y= e x log a

So, \frac{d}{dx} a^x = \frac{d}{dx} e^{x log a}

= \frac{d}{d(x log a)} e^{x log a} \times \frac{d}{dx } x log a

= e^{x log a} \times log a

= a^x \times log a

Thus the derivative of exponential function is found to be:

\frac{d}{dx} a^x = a^x \times log a ]]> http://oscience.info/mathematics/derivatives-of-logarithmic-and-exponential-functions/feed/ 0 Derivatives of inverse trigonometric functions http://oscience.info/mathematics/derivatives-of-inverse-trigonometric-functions/ http://oscience.info/mathematics/derivatives-of-inverse-trigonometric-functions/#comments Tue, 04 Jan 2011 13:01:58 +0000 subash http://oscience.info/?p=504 Inverse trigonometric functions  are the  inverse of trigonometric functions .

For example if, y = sinx  then the inverse function of y = sinx is , is denoted by: x=sin-1y and is called inverse sin function.

You should note that: y = \sin ^{-1} x doesn’t means y = \frac{1}{\sin x} instead

“y = sin -1 x” is the inverse function of “x = sin y”

Derivatives of Inverse Trigonometric Functions:

The derivatives of inverse sine , inverse cos , inverse tan , inverse csc , inverse sec , inverse cot functions are given below:

Derivative of inverse sin function: \frac{d}{dx} \sin ^{-1} x = \frac{1}{\sqrt{1-x^2}}

proof:

If , f(x) = y = \sin ^{-1} x then the function is called inverse sin function.

If, y = \sin ^{-1} x then , x = \sin y

now if we differentiate x = sin y with respect to x , using implicit differentiation technique then,

\frac{d}{dx} x = \frac{d}{dx} \sin y

or, 1 = \frac{d}{dy} \sin y \times \frac{d}{dx} y

or, \frac{d}{dx} y = \frac{1}{cos y}

Now using the trigonometric formula, \cos x = \sqrt{1-(sin x)^2}

\frac{d}{dx} y = \frac{1}{\sqrt{1-(sin y)^2}}

Now as , sin y = x

or, \frac{d}{dx} y = \frac{1}{\sqrt{1-(sin y)^2}} = \frac{1}{\sqrt{1-x^2}}

Thus , \frac{d}{dx} \sin ^{-1} x = \frac{1}{\sqrt{1-x^2}}

Derivative of inverse cos function: \frac{d}{dx} \cos ^{-1} x = \frac{-1}{\sqrt{1-x^2}}

proof:

If , f(x) = y = \cos ^{-1} x then the function is called inverse cos function.

And, If, y = \cos ^{-1} x then we can also rewrite is as: x = \cos y

now if we differentiate x = \cos y with respect to x , using implicit differentiation technique then,

\frac{d}{dx} x = \frac{d}{dx} \cos y

or, 1 = \frac{d}{dy} \cos y \times \frac{d}{dx} y

or, \frac{d}{dx} y = \frac{1}{- \sin y} = \frac{-1}{ \sin y}

Now using the trigonometric formula, \sin x = \sqrt{1-(cos x)^2}

\frac{d}{dx} y = \frac{-1}{\sqrt{1-(cos y)^2}}

Now as , cos y = x

or, \frac{d}{dx} y = \frac{-1}{\sqrt{1-(cos y)^2}} = \frac{-1}{\sqrt{1-x^2}}

Thus , \frac{d}{dx} \cos ^{-1} x = \frac{-1}{\sqrt{1-x^2}}

Derivative of inverse tan function: \frac{d}{dx} \tan ^{-1} x = \frac{1}{1+x^2}

proof:

When , y = \tan ^{-1} x then the function “f” or y is called inverse tan function.
and we can also equally re-write above function as: x = \tan y
If we differentiate both L.H.S and R.H.S of the equation x = \tan y with respect
to “y”.
then,

\frac{d}{dy} x = \frac{d}{dy} \tan y

or, \frac{d}{dy} x = \sec ^2 y = 1 + \tan ^2 y = 1+x^2

Now using the concept of differentials we can re write above equation as:

\frac{d}{dx} y = \frac{1}{1+x^2}

Thus , \frac{d}{dx} \tan ^{-1} x = \frac{1}{1+x^2}

Derivative of inverse csc , inverse sec & inverse cot functions:

We can use the similar method we used above to find derivative of inverse sin , cos and tan function to find the derivatives of inverse csc , sec and cot function.
After differentiation we get following result:
Derivative of inverse csc function: \frac{d}{dx} \csc ^{-1} x = \frac{-1}{x \sqrt{x^2-1}}

Derivative of inverse sec function: \frac{d}{dx} \sec ^{-1} x = \frac{1}{x \sqrt{x^2-1}}

Derivative of inverse cot function: \frac{d}{dx} \csc ^{-1} x = \frac{-1}{1+x^2}

]]> http://oscience.info/mathematics/derivatives-of-inverse-trigonometric-functions/feed/ 0 Derivatives of Trigonometric functions. http://oscience.info/mathematics/derivatives-of-trigonometric-functions/ http://oscience.info/mathematics/derivatives-of-trigonometric-functions/#comments Wed, 08 Dec 2010 08:49:23 +0000 subash http://oscience.info/?p=379 As you know,

The functions SINE x(sin x) , CO-SECANT x(cos x) , TANGENT x(tan x), CO-SECANT x(csc x), SECANT x(sec x) and COTANGENT x(cot x)  are called trigonometrical functions.

You can learn more about these functions by searching about it in the search box above.

We are going to learn and prove ,what are the Derivatives of these Trigonometrical Functions , here.

a> Derivative of sin x:

The derivative of sin x is:

\dfrac{d}{dx}\sin x = \cos x

Proof:

Ley y=SIN x and let this be equation (i)

and let  \Delta x be a small increment in x and \Delta y be the corresponding small increment in y,

Then we can write:

y + \Delta y = \sin (x+ \Delta x) and let it be equation (ii).

Now if we subtract equation (i) from equation (ii), we can get:

\Delta y = \sin (x+ \Delta x) - \sin x

Now using the trigonometrical formula ( sin c – sin d = 2.sin((c-d)/2).cos((c+d)/2)

We get:

\Delta y = 2 \times \sin \frac{\Delta x}{2} \times \cos \frac{2x+\Delta x}{2}

Now dividing both side of above equation by \Delta x we get:

\dfrac{\Delta y}{\Delta x} = \dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\times\cos\frac{2x+\Delta x}{2}

Thus:

\dfrac{d}{dx}y = \dfrac{d}{dx}\sin x = \displaystyle\lim_{{\Delta x}\to 0}\left(\dfrac{\sin \frac{\Delta x}{2}}{\frac{\Delta x}{2}}\times\cos\frac{2x+\Delta x}{2}\right)

or, \dfrac{d}{dx}\sin x = 1 \times \cos \frac{2x}{2} = \cos x

b> Derivative of cos x.

The derivative of cos x is:

\frac{d}{d x}\cos x = -\sin x

Proof:

Let y=cos x

and from trigonometric relations we also know:

\cos x = \sqrt{1-\sin ^{2}x}

So,

\frac{d }{d x} y = \frac{d }{d x}\left(\sqrt{1-\sin ^{2}x}\right)

Now using chain rule :

\frac{d }{d x} y = \frac{d }{d \left(1-\sin ^{2}x\right)} \left(1-\sin ^{2}x\right)^{\frac{1}{2}} \times \frac{d }{d x}\left(1-\sin ^{2}x\right)

or, \frac{d }{d x} y = \frac{0-2\times\sin x\times\cos x}{2\times \sqrt{1-\sin ^{2}x}} = -\sin x

Thus , \frac{d }{d x} \cos x = -\sin x

c> Derivative of  tan x.

The derivative of  tan x is:

\dfrac{d}{dx}\tan x = \sec ^{2}x

Proof:

\dfrac{d}{dx}\tan x = \dfrac{\sin x}{\cos x}

Now using the quotient rule :

or,
\dfrac{d }{dx}\tan x =\dfrac{\cos ^2 x + \sin ^2 x}{\cos ^{2}x}

Thus,

\dfrac{d}{dx}\tan x = \sec ^{2}x

d> Derivative of csc x , secx and cot x.

Using the relation:

csc x= 1/sin x

sec x= 1/cos x

and

cot x = 1/tan x

and then using the quotient rule, we can find the:

Derivative of csc x:

\frac{d }{d x}\csc x = - \csc x \times \cot x

Derivative of sec x:

\frac{d }{d x}\sec x = \sec x \times \tan x

Derivative of cot x:

\frac{d }{d x}\cot x = -\csc ^{2} x ]]> http://oscience.info/mathematics/derivatives-of-trigonometric-functions/feed/ 0 Infinitesimals and Differentials. http://oscience.info/mathematics/infinitesimals-and-differentials/ http://oscience.info/mathematics/infinitesimals-and-differentials/#comments Tue, 23 Nov 2010 16:57:20 +0000 subash http://oscience.info/?p=373 In the chain rule we have come across the following relation:

\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}

Which gives us impression that \frac{dy}{dx} is a fraction of “dy” and “dx”

And in right hand side of above relation we can cancel “du” in the numerator and denominator to get \frac{dy}{dx} of the left hand side.

Actually in the early days Leibnitz and others treated \frac{dy}{dx} as a fraction of “dy” and “dx” or delta y and delta x which are

called infinitesimals or infinitely small numbers having basic properties of the number zero.

In fact this is the reason why British were one or two hundreds years behind than Europe in development of calculus , Europeans regarded dy and dx as infinitesimals

but Britishers didn’t.

As long as a infinitesimals is added or multiplied with a number the infinitesimals acts as the number zero. But unlike zero when an infinitesimals is divided by

another infinitesimals the result is a derivative which can be a function or a constant. This interpretation of \frac{dy}{dx} being fallacious

, Cauchy gave a different type of interpretation and introduced the concept of an operator.

But there is some advantage in considering \frac{dy}{dx} as a fraction , So latter the concept of differential is introduced to replace the idea

infinitesimal.

Let us understand the concept of Differentials with the help of following image:

Concept of Differentials.

In the figure , Let AB be a curve given by the function y=f(x) and P be a point on the curve whose co-ordinates are (x,y) let Q be a point which is very near to P.

So , it’s co-ordinates can be written as:

(x+\Delta x ,y+ \Delta y). Now draw the tangent PT at P which cuts the ordinate QM at T. let PR be perpendicular to QM.

We know that  f ‘ (x) or the derivative of function at x gives the gradient of the curve at the point P or x. which is also measured by:

\dfrac{TR}{PR}

So ,

\dfrac{TR}{PR} = f ‘(x)

But , PR = LM  = OM-OL=   x+\Delta x - x = \Delta x

SO , TR= f ‘(x) * \Delta x

When a point moves along the curve from P to Q , there is change in \Delta x in the abscissa and

\Delta y   in the ordinate of P. But if the point moves along the tangent from P to T , we have change TR in the ordinate for the same change \Delta x in the abscissa.  This change or increment TR is called the differential  of y and is denoted by “dy” or “d f(x)”

So , dy = f ‘ (x)  . \Delta x

The corresponding change or increment  in abscissa which is PR is called the differential of x and is denoted by dx. but PR = \Delta x

So ,

dx = \Delta x

and ,

dy = f ‘ (x)  . \Delta x = f ‘ (x)  .  dx

]]> http://oscience.info/mathematics/infinitesimals-and-differentials/feed/ 0 Derivative of implicit functions. http://oscience.info/mathematics/derivative-of-implicit-functions/ http://oscience.info/mathematics/derivative-of-implicit-functions/#comments Fri, 10 Sep 2010 07:03:36 +0000 subash http://oscience.info/?p=301 explicit functions is easy using the differentiation techniques but It is difficult to find the derivative of implicit functions so we use the Implicit Differentiation technique to find the derivative of Implicit functions. The Implicit Differentiation technique make use of the Chain Rule and the Sum Rule.]]> Finding derivative of a explicit functions is easy using the differentiation techniques but
It is difficult to find the derivative of implicit  functions so we use the Implicit Differentiation technique to find the derivative of Implicit functions.

The Implicit Differentiation technique make use of the Chain Rule and the Sum Rule.

Example of Implicit Differentiation:

let us Differentiate y with respect to x in this Implicit function:

4x3-3y2=21

Differentiating both side of equation with respect to x , and using The Sum Rule:

\dfrac{d}{dx}\left(4x^3\right)-\dfrac{d}{dx}\left(3y^2\right) = \dfrac{d}{dx}\left(21\right)

Now using the The Chain Rule in the second term of the equation:

or,
12x^2-3 \times \dfrac{d}{dy}\left(y^2\right) \times \dfrac{d}{dx}\left(y\right) = o

or,
12x^2-6y \times \dfrac{d}{dx}\left(y\right) = o

SO,

\dfrac{d}{dx}\left(y\right)= \dfrac{12x^2}{6y}

]]> http://oscience.info/mathematics/derivative-of-implicit-functions/feed/ 0 Second and higher derivatives. http://oscience.info/mathematics/second-and-higher-derivatives/ http://oscience.info/mathematics/second-and-higher-derivatives/#comments Fri, 10 Sep 2010 06:07:02 +0000 subash http://oscience.info/?p=291 function y=f(x) , and let y=f(x) be a differentiable function. Then you can differentiate the function f(x) with respect to x or find derivative of f(x) which is called the first derivative of the function f(x). Let the first derivative of the function be "y" and now if you again differentiate y with respect to "x" then the result is called second derivative of function f(x) you can again find it's derivative and that's called the third derivative of function f(x) and similarly you can differentiate again to find fourth , fifth , sixth...... derivative.]]> Think of a function y=f(x) , and let y=f(x) be a differentiable function.

Then you can differentiate the function f(x) with respect to x or find derivative of f(x)= \frac{d}{dx}f(x) which is called the first derivative of the function f(x). Let the first derivative of the function be “y” and now if you again differentiate y with respect to “x” then the result is called second derivative of function f(x) you can again find it’s derivative and that’s called the third derivative of function f(x) and similarly you can differentiate again to find fourth , fifth , sixth…… derivative.

Mathematically you can denote second derivative as:
\frac{d^2}{dx^2}\left(f(x)\right)

The third derivative is denoted by:

\frac{d^3}{dx^3}\left(f(x)\right)

And you can also denote fourth , fifth and all higher derivative by changing the power of “d” and “x” to the respective number.

You can find the second , third , fourth and higher derivative using the following formula:

\frac{d^2}{dx^2}\left(f(x)\right)=\frac{d}{dx}\left(\frac{d}{dx}f(x)\right)

\frac{d^3}{dx^3}\left(f(x)\right)=\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}f(x)\right)\right)
And so on…….

Example of Second and Higher Derivative:

Let us find the second , third , fourth and higher derivatives of the function:

f(x)=5x4+4x3-18

Solution:

The first derivative of the function=
\frac{d}{dx}\left(5x^4+4x^3-18\right) = 20x^3+12x^2

The Second derivative of the function=
\frac{d^2}{dx^2}\left(5x^4+4x^3-18\right) = \frac{d}{dx}\left(20x^3+12x^2\right)=60x^2+24x

The third derivative of the function=
\frac{d^3}{dx^3}\left(5x^4+4x^3-18\right) = \frac{d}{dx}\left(60x^2+24x\right)=120x+24

The fourth derivative of the function=
\frac{d^4}{dx^4}\left(5x^4+4x^3-18\right) = \frac{d}{dx}\left(120x+24\right)=120

The fifth derivative of the function=
\frac{d^5}{dx^5}\left(5x^4+4x^3-18\right) = \frac{d}{dx}120=0

And as fifth derivative of the function f(x) is 0 sixth , seventh , eighth…. and all the higher derivative of the function will be 0.

]]> http://oscience.info/mathematics/second-and-higher-derivatives/feed/ 0 The Chain Rule. http://oscience.info/mathematics/the-chain-rule/ http://oscience.info/mathematics/the-chain-rule/#comments Wed, 08 Sep 2010 12:32:54 +0000 subash http://oscience.info/?p=277 Chain Rule is one of the Techniques of Differentiation.


The Chain Rule states that:

If v(u) is a Function of “u” and u(x) is a function of “x” then  the Derivative of

function “v” with respect to “x” exists which  is equal to the product of derivative of function “v” with respect “u” and derivative of function “u” with respect to “x”.


Mathematically it can be written as:

If,  “v” is a function of “u” and “u” is a function of “x”,


\frac{d}{dx}\left(v\right) = \frac{d}{du}\left(v\right) \times \frac{d}{dx}\left(u\right)





Proof of The Chain Rule:


When, “u” is a function of “x” or u=g(x) and “v” is a function of “u” or v=f(u)

Let, \Delta x be a small increment in “x”. and \Delta u Be the corresponding small increment in “u”.


So,
u+ \Delta u =g(x+\Delta x)


or,
\Delta u=g(x+\Delta x)-g(x)

So,
\displaystyle \lim_{ \Delta x \to o} \Delta u = [g(x+ \Delta x)-g(x)]


=g(x)-g(x)=0

So, \Delta u\rightarrow o as \Delta x\rightarrow o. Then,


We know,
\frac{\Delta v}{\Delta x}=\frac{\Delta v}{\Delta u} \times \frac{\Delta u}{\Delta x}


So,
\frac{d}{dx}\left(v\right) = \displaystyle \lim_{\Delta x\to o}\frac{\Delta v}{\Delta x}=\displaystyle \lim_{\Delta x\to o}\frac{\Delta y . \Delta u}{\Delta u . \Delta x}


=\left(\displaystyle\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta u}\right).\left(\displaystyle\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\right)




Use of Chain Rule:

Find \frac{d}{dx}\left(Y\right) if, y=4u^2-3u+5 and u=2x^2-3


Solution:


\frac{d}{du}\left(Y\right)=8u-3 and \frac{d}{dx}\left(u\right)=4x


So,
\frac{d}{dx}\left(Y\right)=\frac{d}{du}\left(Y\right) . \frac{d}{dx}\left(u\right)


= (8u-3) 4x = [8(2x^2-3)-3] 4x = 64x^3-108x



]]> http://oscience.info/mathematics/the-chain-rule/feed/ 0 The Quotient Rule. http://oscience.info/mathematics/the-quotient-rule/ http://oscience.info/mathematics/the-quotient-rule/#comments Wed, 08 Sep 2010 11:17:07 +0000 subash http://oscience.info/?p=264 Quotient Rule is one of the Techniques of Differentiation.


The Quotient Rule states that:

The Derivative of  a Function “f(x)” divided by another function “g(x)” is the difference between the second function multiplied by derivative of first function and first function multiplied by derivative of second function whole divided by square of  the second function.

Mathematically we can write:

\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\textstyle \frac{g(x).f^|(x)-f(x).g^|(x)}{(g(x))^2}



Proof of Quotient Rule:

If, h(x)=f(x)/g(x)
and “a” is a fixed point then,


h^|(a)=\displaystyle\lim_{x\to a}\frac{h(x)-h(a)}{x-a}


=\displaystyle\lim_{x\to a}\frac{\frac{f(x)}{g(x)}-\frac{f(a)}{g(a)}}{x-a}


=\displaystyle\lim_{x\to a}\frac{g(a)f(x)-f(a)g(x)}{(x-a)g(x)g(a)}


=\displaystyle\lim_{x\to a}\frac{g(a)f(x)-g(a)f(a)+g(a)f(a)-f(a)g(x)}{(x-a)g(x)g(a)}


=
\displaystyle\lim_{x\to a}\left(\frac{1}{g(a).g(x)}\left(g(a).\frac{f(x)-f(a)}{x-a}-f(a).\frac{g(x)-g(a)}{x-a}\right)\right)


=\frac{1}{g(a).g(a)}[g(a).f^|(a)-f(a).g^|(a)]


=\textstyle \frac{g(a).f^|(a)-f(a).g^|(a)}{(g(a))^2}



Use of Quotient Rule:


Find the derivative of: \frac{4x^2+3}{3x^2-2}


Solution:
Let f(x)=4x^2+3 and g(x)=3x^2-2
Then,
\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x).f^|(x)-f(x).g^|(x)}{(g(x))^2}


=\frac{(3x^2-2).8x-(4x^2+3).6x}{(3x^2-2)^2}


=\frac{-34x}{(3x^2-2)^2}

]]> http://oscience.info/mathematics/the-quotient-rule/feed/ 0 The Power Rule. http://oscience.info/mathematics/the-power-rule/ http://oscience.info/mathematics/the-power-rule/#comments Wed, 08 Sep 2010 09:54:44 +0000 subash http://oscience.info/?p=250 Power Rule is one of the Techniques of Differentiation.

The Power Rule states that:

The Derivative of  a Function raised to n’th power is the Product of n , the function raised to the power (n-1)’th and derivative of the function raised to first power.

Mathematically we can write:

\frac{d}{dx}\left(f(x)\right)^n = n\times (f(x))^{(n-1)} \times \frac{d}{dx}\left(f(x)\right)


We can also further synthesize this relation and prove the rule for Derivative of general Polynomial Function. which states:


\frac{d}{dx}\left(x\right)^n = n\times x^{(n-1)}




Proof Of Power Rule for Natural N’th Power:


The power rule is valid for any rational “n”(eg: 4/5) power of a function. But we will prove
the power rule for only natural number power “n” here.


If “f” is a function of “x” then,
By product rule:


\frac{d}{dx}\left(f(x)\right)^2=\frac{d}{dx}\left(f(x).f(x)\right)


= f(x). \frac{d}{dx}\left(f(x)\right)+f(x). \frac{d}{dx}\left(f(x)\right)=2u\frac{d}{dx}\left(f(x)\right)

And,
\frac{d}{dx}\left(f(x)\right)^2=\frac{d}{dx}\left(f(x)^2.f(x)\right)


=f(x).2f(x).\frac{d}{dx}\left(f(x)\right)+f(x)^2\frac{d}{dx}\left(f(x)\right)


=3.f(x).\frac{d}{dx}\left(f(x)\right)


Similarly we can find derivative of fourth power , fifth power………. and the same result comes which is:
\frac{d}{dx}\left(f(x)\right)^n = n\times (f(x))^{(n-1)} \times \frac{d}{dx}\left(f(x)\right)




Use of Power Rule:

Find the derivative of: (7x^3-4)^{3/2}

Solution:

\frac{d}{dx}\left(7x^3-4\right)^{3/2} = \frac{3}{2}\times (7x^3-4)^{1/2}\times \frac{d}{dx}\left(7x^3-4\right)
=\frac{63}{2} \dfrac{1}{2} x^2. \sqrt{7x^3-4}

]]> http://oscience.info/mathematics/the-power-rule/feed/ 0 The Product Rule. http://oscience.info/mathematics/the-product-rule/ http://oscience.info/mathematics/the-product-rule/#comments Wed, 08 Sep 2010 08:51:11 +0000 subash http://oscience.info/?p=237 Product Rule is one of the Techniques of Differentiation.

The Product Rule states that:

The Derivative of  product of two Function is the Sum of Derivative of first function multiplied by second function and first function multiplied by derivative of second function.

Mathematically we can write:

\frac{d}{dx}\left(f(x) \times g(x)\right)=g(x) \times \frac{d}{dx}\left(f(x)\right) +f(x) \times \frac{d}{dx}\left(g(x)\right)

Proof of Product Rule:

If,

h(x)=f(x)\times g(x)

and ‘a’ is a fixed point Then,

h^|(a)=\displaystyle\lim_{x\to a}\frac{h(x)-h(a)}{x-a} =\displaystyle\lim_{x\to a}\frac{f(x).g(x)-f(a).g(a)}{x-a} =\displaystyle\lim_{x\to a}\frac{f(x).g(x)-f(x).g(a)+f(x).g(a)-f(a).g(a)}{x-a} =\displaystyle\lim_{x\to a}f(x)\frac{g(x)-g(a)}{x-a}+\displaystyle\lim_{x\to a}g(a)\frac{f(x)-f(a)}{x-a} =f(a).g^|(a)+g(a).f^|(a)

Use of Product Rule:

Find the derivative of: (3x^2-5x)\times (2x+3)
Solution:

\frac{d}{dx}\left((3x^2-5x)\times (2x+3)\right) = =(2x+3)\frac{d}{dx}\left(3x^2-5x\right )+(3x^2-5x)\frac{d}{dx}\left(2x+3\right ) =(2x+3).(6x-5)+(3x^2-5x).2 =18x^2+8x-25 ]]> http://oscience.info/mathematics/the-product-rule/feed/ 0